Question

Two tiny spheres carrying charges $1.5\mu C$ and $2.5\mu C$ are located 30 cm apart. Find the potential and electric field:

(a) at the mid-point of the line joining the two charges, and

(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Answer 1

Two charges placed at points A and B are represented in the given figure. O is the mid-point of the line joining the two charges.

Magnitude of charge located at A, $q_1=1.5\mu C$

Magnitude of charge located at B, $q_2=2.5\mu C$

Distance between the two charges, $d=30cm=0.3m$

(a)

Let $V_1$ and $E_1$ are the electric potential and electric field respectively at O.

$V_1=$ Potential due to charge at A + Potential due to charge at B

$V_1\frac{q_1}{4\pi {\in }_0\left(\frac{d}{2}\right)}+\frac{q_2}{4\pi {\in }_0\left(\frac{d}{2}\right)}=\frac{1}{4\pi {\in }_0\left(\frac{d}{2}\right)}(q_1+q_2)$

Where,

${\in }_0=$ Permittivity of free space

$\frac{1}{4\pi {\in }_0}=9\times {10}^{9}N{C}^2{m}^{-2}$

$\therefore V_1=\frac{9\times {10}^9\times {10}^{-6}}{\left(\frac{0.30}{2}\right)}$

$E_1=$ Electric field due to $q_2$− Electric field due to $q_1$

$=\frac{q_2}{4\pi {\in }_0{\left(\frac{d}{2}\right)}^2}-\frac{q_1}{4\pi {\in }_0{\left(\frac{d}{2}\right)}^2}$

$=\frac{9\times {10}^9}{{\left(\frac{0.30}{2}\right)}^2}\times {10}^6\times (2.5-1.5)$

$=4\times {10}^{5}V{m}^{-1}$

Therefore, the potential at mid-point is $2.4\times 105V$ and the electric field at mid-point is $4\times 105V{m}^{-1}$. The field is directed from the larger charge to the smaller charge.

(b) Consider a point Z such that normal distance $OZ=10cm=0.1m$, as shown in the following figure.

$V_2$ and $E_2$ are the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

$BZ=AZ=\sqrt{(0.1{)}^2+(0.15{)}^2}=0.18am$

$V_2=$ Electric potential due to A + Electric Potential due to B

$=\frac{q_1}{4\pi {\in }_0\left(AZ\right)}+\frac{q_2}{4\pi {\in }_0\left(BZ\right)}$

$=\frac{9\times {10}^9\times {10}^{-6}}{0.18}(1.5+2.5)$

$=2\times {10}^{5}V$

Electric field due to q at Z,

$E_A=\frac{q_1}{4\pi {\in }_0(AZ{)}^2}$

$=\frac{9\times {10}^9\times 1.5\times {10}^{-6}}{(0.18{)}^2}$

$=0.416\times {10}^{6}V/m$

Electric field due to $q_2$ at Z,

$E_B=\frac{q_2}{4\pi {\in }_0(BZ{)}^2}$

$=\frac{9\times {10}^9\times 2.5\times {10}^{-6}}{(0.18{)}^2}$

$=0.69\times {10}^{6}V{m}^{-1}$

The resultant field intensity at Z,

$E=\sqrt{E_A^2+E_B^2+2E_A{E}_{B}cos2\theta }$

Where, 2θis the angle, $\angle AZB$

From the figure, we obtain

$cos\theta =\frac{0.10}{0.18}=\frac{5}{9}=0.5556$

$\theta =co{s}^{-10}0.5556=56.25$

$\therefore 2\theta ={112.5}^0$

$cos2\theta =-0.38$

$E=\sqrt{(0.416\times {10}^6{)}^2\times (0.69\times {10}^6{)}^2+2\times 0.416\times 0.69\times {10}^{12}\times (-0.38)}$

$=6.6\times {10}^{5}V{m}^{-1}$

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is $2.0\times 1.5V$ and electric field is $6.6\times {10}^{5}V{m}^{-1}$.