Question

slader) Two charges are arranged at corners of a square which has a side length of L = 0.25 m. The values of the charges are q1 = −3.3 × 10−6 C and q2 = +4.2 × 10−6 C. (a) Find the electric potential at points A and B

Answer 1

Explanation:

Formula to calculate the electric potential is as follows.

            $V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}$

Putting the given values into the above formula as follows.

       $V_{A} = \frac{kq_{1}}{\sqrt{2}L} + \frac{kq_{2}}{L}$

               = $\frac{9 \times 10^{9}}{0.25}[\frac{-3.3}{\sqrt{2}} + 4.2] \times 10^{-6}$

               = $6.72 \times 10^{4} V$

Hence, electric potential at point A is $6.72 \times 10^{4} V$.

Now, the electric potential at point B is as follows.

         $V_{B} = \frac{kq_{1}}{L} + \frac{kq_{2}}{\sqrt{2}L}$

                  = $\frac{9 \times 10^{9}}{0.25} [-3.3 + \frac{4.2}{\sqrt{2}}] \times 10^{-6}$

                  = $-1.19 \times 10^{4} V$

Hence, electric potential at point B is $-1.19 \times 10^{4} V$.