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Electric Potential And Potential Energy Due To Point Charges

Practice electric potential due to a point charge and between two charges. Learn electric potential & potential energy definition, examples, formula.

Electric Potential due to a Point Charge

  • The effect of a charge can be defined both by a vector quantity, electric field \((\vec E)\) and by a scalar quantity, electric potential (V).
  • Consider a point P situated at a distance 'r' from a point charge 'q'.

  • The electric potential at point P due to this charge is defined as,

                                            \(V_P=\dfrac {1}{4\pi\epsilon_0}\dfrac {q}{r}\)

NOTE: Put the value of charge with its sign [ (+) positive  or (–) negative].

S.I. Unit:

Unit \(\rightarrow\) Nm/C or Volt

case 1 : when charge is positive 

  • The electric potential at point P due to charge q is defined as, \(V_P=\dfrac {1}{4\pi\epsilon_0}\dfrac {q}{r}\)
  • If charge q is positive, potential V is also positive.
  • Since, \(V\propto\dfrac{1}{r}\) 
  • Thus, positive value of potential will decrease with increase in distance from the charge.
  • Hence, potential will decrease with increase in distance.

case 2 : when charge is negative

  • If charge q is negative, potential V is also negative.
  • Since, \(V\propto-\dfrac{1}{r}\) 
  • Hence, potential will increase with increase in distance.

Illustration Questions

Consider a positive and a negative charge, as shown in figure. Choose the correct option regarding electric potential at given points.

A (VA = VB) > (VC = VD) > (VE = VF)  and (VP = VQ) > (VR = VS) > (VU = VT)

B (VA = VB) < (VC = VD) < (VE = VF)  and (VP = VQ) < (VR = VS) < (VU = VT)  

C (VA = VB) < (VC = VD) < (VE = VF)  and (VP = VQ) > (VR = VS) > (VU = VT)  

D (VA = VB) > (VC = VD) > (VE = VF)  and (VP = VQ) < (VR = VS) < (VU = VT)

×

The value of electric potential decreases with increase in distance from positive charge.

So, (VA = VB) > (VC = VD) > (VE = VF

The value of potential increases with increase in distance from negative charge.

So, (VP = VQ) < (VR = VS) < (VU = VT)

Consider a positive and a negative charge, as shown in figure. Choose the correct option regarding electric potential at given points.

image
A

(VA = VB) > (VC = VD) > (VE = VF)  and (VP = VQ) > (VR = VS) > (VU = VT)

.

B

(VA = VB) < (VC = VD) < (VE = VF)  and (VP = VQ) < (VR = VS) < (VU = VT)  

C

(VA = VB) < (VC = VD) < (VE = VF)  and (VP = VQ) > (VR = VS) > (VU = VT)  

D

(VA = VB) > (VC = VD) > (VE = VF)  and (VP = VQ) < (VR = VS) < (VU = VT)

Option D is Correct

Calculation of Potential

  • Consider a point charge q as shown in figure. To calculate electric potential at different points P, S and R which are r1, r2 and r3 distance apart from the point charge respectively, we use :-

 \(V=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r}\)

So,    \(V_P=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r_1}\)

         \(V_S=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r_2}\)

        \(V_R=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r_3}\)

 

 

Illustration Questions

Calculate the value of potential of a point charge of \(q\) = –\(4\,\mu C\) at a point which is r = 2 m away from the charge. \(\left [ \dfrac {1}{4\pi\epsilon_0}\;\text =\,9\times10^9\,{Nm}^2/\text{C}^2 \right]\)

A – 6 kV

B – 18 kV

C – 13 kV

D – 2 kV

×

Electric potential of a charge at a distance r is given by

 \(V=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r}\)

\(V=\dfrac {1}{4\pi\epsilon_0} \times\;\dfrac {(-4)\times10^{-6}}{2}\)

\(V=9\times10^9\times\dfrac {(-4)}{2}\times10^{-6}\)

\(V = \,– 18\,kV \)

Calculate the value of potential of a point charge of \(q\) = –\(4\,\mu C\) at a point which is r = 2 m away from the charge. \(\left [ \dfrac {1}{4\pi\epsilon_0}\;\text =\,9\times10^9\,{Nm}^2/\text{C}^2 \right]\)

A

– 6 kV

.

B

– 18 kV

C

– 13 kV

D

– 2 kV

Option B is Correct

Potential due to Two Point Charges at any Point

  • Consider two point charges q1 and q2 , as shown in figure. Calculate potential at point P due to two point charges q1 and q2.

Potential of q1 at P, \(V_{P_1}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_1}{r_1}\)

and 

Potential of q2 at P, \(V_{P_2}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_2}{r_2}\)

  • Since potential is a scalar quantity. So, total electric potential at P will be scalar sum of potential of q1 and q2 at P.

\(V_P=V_{P_1}+V_{P_2}\)

Flow Chart Representation for Calculation of Potential

Illustration Questions

Calculate the total electric potential at a point P, situated at a distance r1 = 2 m from charge \(q_1=-3\,\mu C \) and r2 = 3 m from charge \(q_2=2\,\mu C\).

A –7.5 kV

B 8.9 kV

C –9.8 kV

D 10 kV

×

Electric potential of charge q at r distance

\(V\)\(\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r}\)

 

image

Potential of  \(-3\,\mu C\) at point P (V1)

\(V_1=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {(-3)\times10^{-6}}{2}\)

\(V_1= 9\times10^9 \times \dfrac {(-3)}{2}\times10^{-6}\)

\(V_1 = –13.5 × 10^3 \,V\)

\(V_1 = –13.5 \,kV\)

image

Potential of \(2\,\mu C\) at point P (V2)

\(V_2=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {2\times10^{-6}}{3}\)

\(V_2= 9\times10^9 \times \dfrac {2}{3}\times10^{-6}\)

\(V_2 = 6 × 10^3 \,V\)

\(V_2 = 6 \,kV\)

image

Total electric potential at point P (V)

\(V = V_1 + V_2\)

\(V = – 13.5\,kV + 6 \,kV\)

\(V = –7.5 \,kV\)

image

Calculate the total electric potential at a point P, situated at a distance r1 = 2 m from charge \(q_1=-3\,\mu C \) and r2 = 3 m from charge \(q_2=2\,\mu C\).

image
A

–7.5 kV

.

B

8.9 kV

C

–9.8 kV

D

10 kV

Option A is Correct

Electric Potential at the Center of Symmetric Charge Distribution (Sum Zero)

  • Consider a symmetric charge distribution of four charges +q, –q, +2q and –2q which are placed on the circumference of a circle of radius 'r', as shown in figure.

  • From figure, it can be concluded that sum of all charges is zero.
  • Total electric potential at the center of circle (O) is the scalar sum of potential due to individual charge.
  • Electric potential of q at O,

 \(V_{q}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q}{r}\)

  • Electric potential of –q at O,

 \(V_{-q}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {(-q)}{r}\)

  • Electric potential of 2q at O,

 \(V_{2q}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {2q}{r}\)

  • Electric potential of –2q at O, 

\(V_{-2q}=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {(-2q)}{r}\)

  • Total electric potential at O,

\(V = V_q + V_{-q} + V_{2q} + V_{-2q}\)

\(V\) \(=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {+q}{r}+ \dfrac {1}{4\pi\epsilon_0}\;\dfrac {(-q)}{r}+ \dfrac {1}{4\pi\epsilon_0}\;\dfrac {2q}{r}+ \dfrac {1}{4\pi\epsilon_0}\;\dfrac {(-2q)}{r} \)

\(V = 0\)

Conclusion: If sum of all charges placed symmetrically is zero, then total electric potential at center of symmetry will also be zero.

 

Illustration Questions

Three charges q, 2q and –3q are placed on the three vertices of an equilateral triangle of sides equal to a, as shown in figure. Find the electric potential at the centroid of the triangle.

A \(\dfrac {4q}{4\pi\epsilon_0a}\)

B \(\dfrac {3q}{4\pi\epsilon_0a}\)

C \(\dfrac {-q}{4\pi\epsilon_0a}\)

D \(Zero\)

×

Sum of all charges = q + 2q –3q = 0

image

Since, sum of charges = 0

So, Electric potential at centroid = 0

image

Three charges q, 2q and –3q are placed on the three vertices of an equilateral triangle of sides equal to a, as shown in figure. Find the electric potential at the centroid of the triangle.

image
A

\(\dfrac {4q}{4\pi\epsilon_0a}\)

.

B

\(\dfrac {3q}{4\pi\epsilon_0a}\)

C

\(\dfrac {-q}{4\pi\epsilon_0a}\)

D

\(Zero\)

Option D is Correct

Electric Potential at the Center of Symmetric Charge Distribution (Sum Non-Zero)

  • Consider a symmetric charge distribution in which four charges are placed at the vertices of a square of side 'a', as shown in figure,

  • From figure, it is concluded that sum of all the four charges is not zero.
  • Total electric potential at the center of square (O) will be scalar sum of potential due to all charges.
  • Electric potential of q at O,

 \(V_{1}=\dfrac {1}{4\pi\epsilon_0}\;\times\dfrac {q}{(a/\sqrt{2})}\)

  • Electric potential of 2q at O, 

\(V_{2}=\dfrac {1}{4\pi\epsilon_0}\;\times\dfrac {2q}{(a/\sqrt{2})}\)

  • Electric potential of –2q at O, 

\(V_{3}=\dfrac {1}{4\pi\epsilon_0}\;\times\dfrac {(-2q)}{(a/\sqrt{2})}\)

  • Electric potential of q at O, 

\(V_{4}=\dfrac {1}{4\pi\epsilon_0}\;\times\dfrac {q}{(a/\sqrt{2})}\)

  • Total electric potential at point O,

\(V = V_1 + V_2 + V_3 + V_4\)

\(V=\dfrac {2\sqrt{2}\;q}{4\pi\epsilon_0\;a}\)

Conclusion: Total electric potential is the scalar sum of all electric potential due to individual charge.

Flow chart:

Illustration Questions

Four charges, each of \(q =\)  \(5\,\mu C\) are placed at four vertices of square of side \(\ell \,= 2\,m \) each. Calculate total electric potential at the center of the square.

A \(30\)\(\sqrt{2}\) \(kV\)

B \(45\)\(\sqrt{2}\) \(kV\)

C \(90\)\(\sqrt{2}\) \(kV\)

D \(60\)\(\sqrt{2}\) \(kV\)

×

Using Pythagoras theorem

(2x)2 = (2)2 + (2)2

4x2 =  2(2)2

\(x=\sqrt 2\,m\)

image

Electric potential at O,

Vo = V1 + V2 + V3 + V4

Vo = 4 V  (All the \(5\mu C\) are placed at equal distance from O)

\(V_O=4\times\dfrac {1}{4\pi\epsilon_0}\times\dfrac {5\times10^{-6}}{\sqrt{2}}\)

\(=\dfrac {4\times9\times10^9\times5\times10^{-6}} {\sqrt{2}}\)

\(=90\sqrt{2}\) \(kV\)

image

Four charges, each of \(q =\)  \(5\,\mu C\) are placed at four vertices of square of side \(\ell \,= 2\,m \) each. Calculate total electric potential at the center of the square.

image
A

\(30\)\(\sqrt{2}\) \(kV\)

.

B

\(45\)\(\sqrt{2}\) \(kV\)

C

\(90\)\(\sqrt{2}\) \(kV\)

D

\(60\)\(\sqrt{2}\) \(kV\)

Option C is Correct

Concept of Potential Energy

Definition- Change in electrical  potential energy of a system is defined as the negative of work done by electrical forces in changing the configuration of the system.

  • Electric potential energy is a scalar quantity.

S.I. Unit:  Joule

  • Consider two positive point charges, q1 and q2, separated with a distance r1, as shown in figure.

Now, q2 is taken to a point C slowly (so that no potential energy changes into kinetic energy), situated at a distance r2 from q1.

  • Work done by electrical forces due to q1 in moving charge q2, from B to C can be calculated as:
  • Consider a point P, between B and C at a distance r from A.
  • Electrical force at P,

    \(\vec F = \dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_1q_2}{r^2}\) (A to C)

  • Work done by electrical force to displace q2 by a small distance dr from P to C, 
  • \(dW=\vec F.d\vec r\)

    \(dW = \dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_1q_2}{r^2}\;dr\)

[\(\vec F \) and \(d \vec r\) are in same direction]  

  • Total work done to displace q2 from B to C,

\(W_{(B\rightarrow C)}=\int\limits_{r_1}^{r_2}\dfrac {1}{4\pi\epsilon_0}\dfrac {q_1q_2}{r^2}dr\)

\(W_{(B\rightarrow C)}=\dfrac {1}{4\pi\epsilon_0}\times q_1q_2 \left [ \dfrac {1}{r_2}-\dfrac {1}{r_1} \right]\)

 \(\left [ \because\int\limits_{r_1}^{r_2}\dfrac {1}{r^2} \;dr= \left [ \dfrac {-1}{r} \right]_{r_1}^{r_2}= \left [ -\dfrac {1}{r_2}+\dfrac {1}{r_1} \right] \right]\) 

  • From definition of electric potential energy, change in potential energy from B to C,

\(\Delta U=U_C-U_B=-W_{B \rightarrow C}\)

\(\Rightarrow\)\(U_C-U_B=-\dfrac {q_1q_2}{4\pi\epsilon_0} \left [ \dfrac {-1}{r_2}+\dfrac {1}{r_1} \right]\)

\(\Rightarrow \)\(U_C-U_B=\dfrac {q_1q_2}{4\pi\epsilon_0} \left [ \dfrac {1}{r_2}-\dfrac {1}{r_1} \right]\)

  • Potential energy of two particle system is considered as zero when they are widely separated.

\(U_{\infty}=0\)

 

  • Potential energy of two particle system when they are at a distance r from each other.

\(\Delta U=U_r-U_{\infty}\)

\(=U_r-U_{\infty}\)

\(=\dfrac {1}{4\pi\epsilon_0}\times q_1q_2 \left[ \dfrac {1}{r}- \dfrac {1}{\infty} \right]\)

\(U_r=\dfrac {q_1q_2}{4\pi\epsilon_0\;r} \)

  • The potential energy of two positive point charges q1 and q2 situated at a distance 'r' is given by

\(U_r=\dfrac {q_1q_2}{4\pi\epsilon_0r} \)

Note: Value of charges are considered with sign.

Illustration Questions

Calculate the potential energy of a system which consists of two charge particles  \(q_1=4\,\mu C\) and \(q_2=-4\,\mu C\) separated by a distance r = 2 m. \(\left [ \dfrac {1}{4\pi\epsilon_0}=9\times10^9\, \text {Nm}^2/\text{C}^2 \right]\)

A –72 m J

B –40 m J

C –36 m J

D –78 m J

×

Potential energy of two charge particles

\(U=\dfrac {1}{4\pi\epsilon_0}\;\dfrac {q_1q_2}{r}\)

image

Put the values, 

q1 = 4 × 10–6 C

q2 = –4 × 10–6 C

and r = 2 m

\(U=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {4\times10^{-6}\times(-4)\times10^{-6}}{2}\)

U = 9 × 109  ×[ (–8) × 10–12 ]

U = – 72 × 10–3 

U = – 72 mJ

image

Calculate the potential energy of a system which consists of two charge particles  \(q_1=4\,\mu C\) and \(q_2=-4\,\mu C\) separated by a distance r = 2 m. \(\left [ \dfrac {1}{4\pi\epsilon_0}=9\times10^9\, \text {Nm}^2/\text{C}^2 \right]\)

image
A

–72 m J

.

B

–40 m J

C

–36 m J

D

–78 m J

Option A is Correct

Potential Energy of a System

  • Consider a system consisting of more than two point charges.
  • Potential energy for a pair of charges,

 \(U=\dfrac {q_1q_2}{4\pi\epsilon_0r}\)

  • As potential energy is a scalar quantity, total potential energy is the scalar sum of all possible pairs presented in the system.
  • For three point charges, three pairs are possible and for four point charges 6 pairs are possible and so on.

Flow Chart:

Illustration Questions

Calculate the total potential energy of the system given below.

A 6.3 J

B 9.2 J

C 3.6 J

D 2.1 J

×

Potential energy due to charges  q1 and q2

\(U_{q_1q_2}=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {q_1q_2}{r_1}\)

\(U_{q_1q_2}=\dfrac {9\times10^9\times(-1)\times10^{-6}\times6\times10^{-6}}{1\times10^{-2}}\)

\(U_{q_1q_2}=\) \(–5.4 \,J\)

image

Potential energy due to charges  q2 and q3

\(U_{q_2q_3}=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {q_2q_3}{r_2}\)

\(U_{q_2q_3}=\dfrac {9\times10^9\times6\times10^{-6}\times4\times10^{-6}}{2\times10^{-2}}\)

\(U_{q_2q_3}=\) \(10.8 \,J\)

image

Potential energy due to charges  q1 and q3

\(U_{q_1q_3}=\dfrac {1}{4\pi\epsilon_0}\times\dfrac {q_1q_3}{r_3}\)

\(U_{q_1q_3}=\dfrac {9\times10^9\times(-1)\times10^{-6}\times4\times10^{-6}}{2\times10^{-2}}\)

\(U_{q_1q_3}=\) \(–1.8 \,J\)

image

Total potential energy of the system

=\(U_{q_1q_2}+U_{q_2q_3}+U_{q_1q_3}\)

\(U = –5.4 + 10. 8 – 1.8 \)

\(U = 3.6 \,J\)

image

Calculate the total potential energy of the system given below.

image
A

6.3 J

.

B

9.2 J

C

3.6 J

D

2.1 J

Option C is Correct

Illustration Questions

Consider a point charge q0, at the center of two concentric circles of radius r1 = R and r2 = 2R. Potential due to charge q0 on a circle of radius R is V1 = 5 V. Calculate the potential due to q0 on a circle of radius 2R.

A 3 V

B 2.5 V

C 8 V

D 5 V

×

Electric potential on a circle of radius R due to q0 is

\(V_1=\dfrac {k\;q}{R}\)

\(5=\dfrac {k\;q_0}{R}\) ...(i)

image

Electric potential on a circle of radius 2R due to q0 is

\(V_2=\dfrac {k\;q}{R}\)

\(V_2=\dfrac {k\;q_0}{2R}=\dfrac {1}{2} \left( \dfrac {k\,q_0}{R} \right)\) ...(ii)

image

From (i) and (ii)

\(V_2=\dfrac {1}{2}(5)\)

\(V_2 = 2.5\, V\)

image

Consider a point charge q0, at the center of two concentric circles of radius r1 = R and r2 = 2R. Potential due to charge q0 on a circle of radius R is V1 = 5 V. Calculate the potential due to q0 on a circle of radius 2R.

image
A

3 V

.

B

2.5 V

C

8 V

D

5 V

Option B is Correct

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