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Calculation Of Potential From Electric Field

Practice to finding potential at any point in a uniform electric field and inside a sphere, spherical shell and a hollow sphere. Learn relation between potential and electric field.

Relation between Potential and Electric Field

  • Electric field and potential are two different parameters to understand the effect of a charge but they are mutually related to each other.
  • Consider a charge +Q. To understand its field and potential, consider a small positive test charge q0, placed initially at a distance ri from +Q.

  • Displacing the charge q0 from ri to r, change in potential energy between these two points is \(\Delta U\).

  • Potential difference between these two points is defined as change in potential energy per unit test charge.

\(\Delta V=\dfrac {\Delta U}{q_0}\)

  • As electric field is conservative, so change in potential energy will be  equal to the negative of work done by electric force.

So,

\(\Delta U=-W_{el}\)

\(\Delta V=\dfrac {\Delta U}{q_0}=\dfrac {-W_{el}}{q_0}\)

\(\Delta V=-\dfrac{\int\limits _{r_i}^{r_{f}} {\vec F.d\vec s}}{q_0}\)         \(\left [ W=\int \limits_{i}^{f} \vec F.d\vec s \right]\)

\(\Delta V=-\dfrac{\int\limits _{r_i}^{r_{f}} {q_0\vec E.d\vec s}}{q_0}\)

\(\Delta V=-\int \limits_{r_i}^{r_{f}} \vec E.d\vec s\)

Conclusion: Potential difference between two points in an electric field is obtained by performing line integral of electric field between those two points.

Case I:

  • Consider a positive charge +Q. A test charge is moved away from this charge as shown in figure.

  • As \(d\vec s\) and \(\vec E\) are in same direction, so \(\vec E.d\vec s\) will be positive.

\(\Delta V=-\int \vec E.d\vec s\)

\(\Delta V=\)Negative

Conclusion: As the distance from the positive charge increases, the potential decreases.

 

Case II:

  • Consider a negative charge –Q. A test charge is moved away from this point as shown in figure.

  • As \(d\vec s\) and \(\vec E\) are in opposite direction, so \(\vec E.d\vec s\) will be negative.
  • \(\Delta V=-\int \vec E.d\vec s\)

    \(\Delta V=\)Positive

    Conclusion: As the distance from the negative charge increases, the potential increases.

  • Conclusively, on moving in the direction of electric field vector, potential decreases and on moving in opposite direction of electric field vector, potential increases.

Illustration Questions

Choose the correct option regarding to the order of values of electric potential at different points.

A

B

C

D

×

As the distance from positive charge increases, potential decreases. So, VA > VB > VC. Option A is incorrect.

As the distance from negative charge increases, potential increases. So, VC > VB > VA. Option B is correct.

Moving in the direction of electric field vector, potential decreases. So, VA > VB > VC. Option C is incorrect.

Moving in the direction of electric field vector, potential decreases. So, VA > VB > VC. Option D is incorrect.

Choose the correct option regarding to the order of values of electric potential at different points.

A image
B image
C image
D image

Option B is Correct

Finding Potential at any Point in a Uniform Electric Field

  • Consider a uniform electric field directed along positive x-axis, as shown in figure.
  • In figure,  \(\vec E=|\vec E| \hat i\)

 

  • Consider a point A, where electric potential is VA and another point B at a distance r, at an angle \(\theta\) above the horizontal.
  • To find electric potential at point B(VB) we use,

\(\Delta V_{AB}=-\int\limits_A^B\vec E . d\vec r\)

\(\Rightarrow V_B-V_A=-\int\limits_A^B\vec E.d \vec r\)

\(\Rightarrow V_B-V_A=-\int\limits_A^B E dr\cos\theta\)

\(\Rightarrow V_B-V_A=-E\cos\theta \int\limits_A^B dr\)

\(\Rightarrow V_B-V_A=-E \cos\theta \,r\)

\(\Rightarrow V_B=V_A-E(r \cos \theta)\)

  • Here, \(r \,cos \,\theta\)  is the component of displacement along the direction of electric field.

Conclusion: In a uniform electric field, potential difference between any two points gets affected only because of component of displacement along the electric field vector.

  • If the component of displacement along the electric field vector is zero, then points are considered to be at same potential, i.e potential difference between them will be zero.

Illustration Questions

Two points A and B, are placed in a uniform electric field of 10 V/m, as shown in figure.If the potential at point A is 15 volt, then calculate potential at point B.

A 20 V

B 50 V

C 30 V

D 10 V

×

To find electric potential at the point B(VB) we use,

\(\Delta V_{AB}=-\int\limits_A^B\vec E . d\vec r\)

\(\Rightarrow V_B-V_A=-\int\limits_A^B\vec E.d \vec r\)

\(\Rightarrow V_B-V_A=-\int\limits_A^B E \,dr\cos\theta\)

\(\Rightarrow V_B-V_A=-E\,\cos\theta \int\limits_A^B dr\)

\(\Rightarrow V_B-V_A=-E\, \cos\theta\, r\)

\(\Rightarrow V_B=V_A-E(r \cos \theta)\)

Here, V= 15 V, r = 1m, \(\theta\)= 60°

V= 15 – 10 (1 cos 60°)

V=10 V

Two points A and B, are placed in a uniform electric field of 10 V/m, as shown in figure.If the potential at point A is 15 volt, then calculate potential at point B.

image
A

20 V

.

B

50 V

C

30 V

D

10 V

Option D is Correct

Calculation of Potential when Electric Field is Given as a Function of x

  • Consider a variable electric field E, directed along x-axis such as,

\(\vec E=f(x) \hat i\)

  • Consider two points A and B, at x-axis such that their position are at x = x1 and x = x2, respectively.

  • Potential difference between A and B,

\(V_B-V_A=-\int\limits_{x_1}^{x_2}\vec E. d\vec x\)

\(V_B-V_A=-\int\limits_{x_1}^{x_2}| \vec E|\; d x\)

  • Since \(\vec E\) and \(d\vec x\) are in same direction. So,

\(\vec E. d \vec x=| \vec E| \;d x\)

\(V_B=V_A-\int\limits_ {x_1}^{x_2}|\vec E|\; dx\)

\(V_B=V_A-\int\limits_ {x_1}^{x_2}f(x) \;dx\)

Illustration Questions

Two points A and B, are situated on positive x-axis in a non-uniform electric field varying as \(\vec E=(\dfrac {5}{x} \hat i) V/m\) such that their positions are xA = 1m and xB = 5m. If the electric potential at point A is 15 Volt then, electric potential at point B will be

A 19.5 Volt

B 11.5 Volt

C 25 Volt

D 100 Volt

×

\(\vec E=\dfrac {5}{x} \hat i V/m\), VA=15 Volt, VB = ?

Potential difference between A and B is given as

\(V_B-V_A=-\int\limits_{x_A}^{x_B} E\; dx\)

\(V_B=V_A-\int\limits_{x_A}^{x_B} E\; dx\)

\(V_B=15-\int\limits_{1}^{5} \dfrac {5}{x} dx\)

\(V_B=15-5 [\; log\, x\; ] _1^5\)

\(V_B=15-5\;log\,5\)

\(V_B=15-5(0.7)\)

\(V_B = 11.5\, Volt\)

Two points A and B, are situated on positive x-axis in a non-uniform electric field varying as \(\vec E=(\dfrac {5}{x} \hat i) V/m\) such that their positions are xA = 1m and xB = 5m. If the electric potential at point A is 15 Volt then, electric potential at point B will be

A

19.5 Volt

.

B

11.5 Volt

C

25 Volt

D

100 Volt

Option B is Correct

Potential at a Point Inside a Hollow Charged Sphere

Electric Field Representation of Spherical Shell

  • Consider a spherical shell of radius R and charge +Q. Electric field due to this charge distribution as shown in figure.

  • To calculate electric potential difference between two points,

\(V_B-V_A = \dfrac { \Delta U_{AB}} {q_0}\)

 

  • As electric force is conservative in nature,
  • So, \(\Delta U_{AB}=-W_{el}\)

                  \(U_B-U_A=-W_{el}\)

  • Potential energy of two charge system when they are infinitely separated, is zero.

\(U_\infty=0\)

  • If point A is considered at infinity, negative of work done by electric force between infinity to that point gives the potential energy at point B.
  • So, potential energy at point B,

\(U_B-U_A=-W_{el\,A\rightarrow B}\)

\(U_B-0=-W_{{el}\,\infty\rightarrow B}\)

  • So, potential at point B,

\(V_B=\dfrac {U_B}{q_0}=\dfrac {-W_{{el}\,\infty \rightarrow B}}{q_0}\)

  • Conclusively, potential at any point is the negative of work done in moving a charge from infinity to that point per unit test charge. So, potential at any point,

\(V=\dfrac {-W_{{el}\,\infty \rightarrow point}}{q_0}\)

\(=\dfrac {-\int\limits_{\infty}^{\ell}\vec F_{el}.d\vec \ell}{q_0}\)

\(V=\dfrac {-\int\limits_{\infty}^{\ell}q_0\;\vec E.d\vec \ell}{q_0}\\\)

\(V= {-\int\limits_{\infty}^{\ell}\;\vec E.d\vec \ell}\)

  • Since \(\vec E\) changes its value from \(\infty\)to \(2\ell\), we break the limits into two parts.

\(V=-\left [ \int\limits_{\infty}^{2\ell} \vec E_1.d\vec \ell + \int\limits_{2\ell}^{\ell} \vec E_2.d\vec \ell \right]\)

  • To find potential at a point P inside a spherical shell at a distance x from center, consider a spherical shell of radius R and charge +Q situated at infinity. 
  • Charge Q is moved from infinity to point P'' outside the sphere.

Potential at Point P"

\(V_{P \;surface}=-\int\limits_{\infty}^{r}\vec E.d\vec\ell=- \left [ \int\limits_{\infty}^{r}\; E_{outside}\;.dr \right]\)

\(V_{P \;surface}=- \left [ \int\limits_{\infty}^{r}\; \dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{r^2} dr \right] =\dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{r}\)

At r = R,  potential at surface = \(\dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{R}\)

  • Since potential at surface of the sphere is constant. Also, the electric field inside a hollow sphere is zero. So, the potential at a point inside a hollow sphere will be same as potential on the surface.

\(V_{inside}=\dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{R}\)

\(\because\vec E=0\)\(V=-\int\vec E.d\vec r\). So, V = constant ]

Illustration Questions

A hollow sphere of radius R = 5 cm and charge Q = +5\(\mu\)C is as shown. Calculate the value of electric potential at point P which is at a distance of x = 2 cm from its center.

A 15 × 106 Volt

B 9 × 105 Volt

C 6 × 1010 Volt

D 7 × 109 Volt

×

Potential at a point inside a hollow sphere = \(\dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{R}\)

image

\(V_P=\dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{R}\)

where,  \(\dfrac {1}{4\pi\epsilon_0} = 9\times10^9\; Nm^2/C^2\)

\(Q = 5 × 10^{-6}\,C, R = 5×10^{-2}\,m, x = 2\,cm\)

image

\(V_P=\dfrac {9\times10^9\times 5\times10^{-6}} {5\times10^{-2}}=9\times10^{5}\)

\(V_P = 9 × 10^5\, Volt\)

image

A hollow sphere of radius R = 5 cm and charge Q = +5\(\mu\)C is as shown. Calculate the value of electric potential at point P which is at a distance of x = 2 cm from its center.

image
A

15 × 106 Volt

.

B

9 × 105 Volt

C

6 × 1010 Volt

D

7 × 109 Volt

Option B is Correct

Potential at a Point

  • Consider a region of a non-uniform electric field such that, 

\(\vec E = E_x \hat i+E_y \hat j\)

  • Electric potential at the origin is V0. To calculate potential at a point P(x, y),

\(V_P-V_0=-\int\limits_0^P\vec E .d\vec \ell\)

\(V_P-V_0=-\int\limits_{(0,0)}^{(x,y)} (E_x\hat i - E_y\hat j).(dx\hat i+dy\hat j)\)

\(V_P-V_0=-\int\limits_{0}^{P} E_x \;dx -\int\limits_{0}^{P} E_y\; dy \)

\(V_P-V_0=-\int\limits_{0}^{x} E_x \;dx -\int\limits_{0}^{y} E_y\; dy \)

\(V_P=V_0-\int\limits_{0}^{x} E_x \;dx -\int\limits_{0}^{y} E_y\; dy \)

Illustration Questions

An electric field in X-Y plane is as \(\vec E=\) \((x^2\;\hat i+2y\;\hat j)\,\)Volt/m. Find the electric potential at a point P(3,4) if electric potential at origin is V=2 Volt.

A –23 Volt

B –45 Volt

C –50 Volt

D –70 Volt

×

Figure representing point P placed in non uniform electric field.

image

\(V_P-V_O=-\int\limits_{0}^{3} E_x \;dx -\int\limits_{0}^{4} E_y\; dy \)

\(V_P-2=-\int\limits_{0}^{3} x^2 \;dx -\int\limits_{0}^{4} 2y\; dy \)

 

image

\(V_P-2=- \left[ \dfrac {x^3}{3}\right]_0^3 -2\left [\dfrac {y^2}{2}\right]_0^4\)

\(V_P-2=- \left( \dfrac {3^3}{3}\right) -2\left (\dfrac {4^2}{2}\right)\)

\(V_P = 2 – 9 – 16\)

\(= – 23\,Volt\)

image

An electric field in X-Y plane is as \(\vec E=\) \((x^2\;\hat i+2y\;\hat j)\,\)Volt/m. Find the electric potential at a point P(3,4) if electric potential at origin is V=2 Volt.

A

–23 Volt

.

B

–45 Volt

C

–50 Volt

D

–70 Volt

Option A is Correct

Potential Inside a Uniformly Charged Sphere

  • Consider a solid sphere of charge +Q distributed uniformly in its volume.
  • The variation of electric field (E) with the distance from the center of the sphere (r) is shown.

  • To calculate potential at a point P inside the sphere at a distance r from its center.

\(V_P=-\int\limits_\infty^P\vec E.d\vec r\)

 

 

Note: [\(\because\) Value of electric field is different inside and outside of solid sphere so, we break limits]

\(V_P=-\left [ \int\limits_\infty^R\vec E_{outside}.d\vec r +\int\limits_R^r\vec E_{inside}.d\vec r \right]\)

\(V_P=-\left [ \int\limits_\infty^R \left\{ \dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{r^2} dr \right\} + \int\limits_R^r \left\{ \dfrac {1}{4\pi\epsilon_0} \dfrac {Qr}{R^3} dr \right\} \right]\)

\(V_P=- \dfrac {Q}{4\pi\epsilon_0} \left [ \left\{ \dfrac {-1}{r}\right\}_{\infty}^R + \dfrac {1}{R^3} \left\{ \dfrac {r^2}{2} \right\}_R^r \right]\)

\(V_P= \dfrac {Q}{4\pi\epsilon_0} \left [\dfrac {1}{R} \right] + \dfrac {Q}{4\pi\epsilon_0} \dfrac {1}{R^3} \left[ \dfrac {R^2}{2}-\dfrac {r^2}{2} \right]\)

\(V_P=\dfrac {Q}{4\pi\epsilon_0R}+\dfrac {Q}{4\pi\epsilon_0(2R)}-\dfrac {Qr^2}{4\pi\epsilon_0(2R^3)}\)

\(V_P=\dfrac {3Q}{2(4\pi\epsilon_0)R}-\dfrac{Q}{4\pi\epsilon_0} \left( \dfrac {r^2}{2R^3} \right)\)

\(V_P=\dfrac {Q}{4\pi\epsilon_0R^3} \left( \dfrac {3}{2}R^2-\dfrac {r^2}{2} \right)\)

\(V_P=\dfrac {Q}{4\pi\epsilon_0R^3} \left( \dfrac {3R^2-r^2}{2} \right)\)

Illustration Questions

Find the electric potential at a point which is at a distance x = 2 cm from the center of a sphere of radius R = 10 cm and charge Q = +5 \(\mu\)C.

A 66 MV

B 666 kV

C 55 kV

D 0 V

×

Potential at a point P inside a uniformly charged sphere is given as

\(V_P=\dfrac {Q}{4\pi\epsilon_0R^3} \left( \dfrac {3R^2-r^2}{2} \right)\)

image

\(V_P=\dfrac {5\times 10^{-6}\times 9\times 10^9}{(10\times 10^{-2})^3} \left( \dfrac {3(10\times 10^{-2})^2-(2\times 10^{-2})^2}{2} \right)\)

\(V_P=45\times10^6\times 10^{-4} \left( \dfrac {296}{2} \right)\)

\(V_P=6660\times10^2\) \(Volt\)

\(V_P=666\, \text{kV}\)

image

Find the electric potential at a point which is at a distance x = 2 cm from the center of a sphere of radius R = 10 cm and charge Q = +5 \(\mu\)C.

A

66 MV

.

B

666 kV

C

55 kV

D

0 V

Option B is Correct

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