Informative line

### 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.

#### 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
B
C
D

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.

#### 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.

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,

$$\vec E. d \vec x=| \vec E| \;d x$$

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

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

#### 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 ]

#### 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 = $$\dfrac {1}{4\pi\epsilon_0} \dfrac {Q}{R}$$

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

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

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

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

$$V_P = 9 × 10^5\, Volt$$

### 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

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),

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

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

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

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

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

#### 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.

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

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

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

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

$$V_P = 2 – 9 – 16$$

$$= – 23\,Volt$$

### 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]

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

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

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

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

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

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

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

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

#### 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

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

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

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

$$V_P=6660\times10^2$$ $$Volt$$

$$V_P=666\, \text{kV}$$

### 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