Question

A charge $q_1=2\mu C$ is located at the origin and a charge $q_2=-6\mu C$ is located at $(0,3m)$. Find the total electric potential due to the charges at point $(4m,0)$

• A. $-6.3eV$
• B. $-6.3KV$
• C. $-6.3MV$
• D. $-6.3V$

Answer 1

The correct option is B $-6.3KV$

$\text{Step 1: Potential due to}$ $q_1$ $\text{to}$ $q_2$ $\text{[Ref. Fig.]}$

From Figure distance $r_2=5m$

At point $A$

Potential due to $q_1$, $V_1=\frac{K{q}_1}{r_1}=\frac{9\times {10}^9\times (2\times {10}^{-6}C)}{4m}$

$\Rightarrow V_1=4.5\times {10}^{3}Volts$

Potential due to $q_2$, $V_2=\frac{K{q}_2}{r_2}=\frac{9\times {10}^9\times (-6\times {10}^{-6}C)}{5m}$

$\Rightarrow V_2=-10.8\times {10}^{3}Volts$

$\text{Step 2: Net Potential at Point (A)}$

Potential is a scalar quantity

So, $V_{net}=V_1+V_2$

$=4.5\times {10}^3-10.8\times {10}^3=-6.3\times {10}^{3}volts$

$V_{net}=-6.3KV$

Hence, option $B$ is correct.