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.

- 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 q
_{0,}placed initially at a distance r_{i}from +Q.

- Displacing the charge q
_{0}from r_{i}to r_{f }, 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.

- 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.

- 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.

- 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 V
_{A}and another point B at a distance r, at an angle \(\theta\) above the horizontal. - To find electric potential at point B(V
_{B}) 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.

- 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 V
_{0}. To calculate potential at a point P(x, y),

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

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

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

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

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

- 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**]**

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

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

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

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

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

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

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

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

- 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 = x
_{1}and x = x_{2,}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,

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

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

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

A 19.5 Volt

B 11.5 Volt

C 25 Volt

D 100 Volt

- 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.

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

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

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 15 × 106 Volt

B 9 × 105 Volt

C 6 × 1010 Volt

D 7 × 109 Volt