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

Step-by-step explanation:

Formula to calculate the electric potential is as follows.

`V_(A) = (kq_(1))/(√(2)L) + (kq_(2))/(L)`

Putting the given values into the above formula as follows.

`V_(A) = (kq_(1))/(√(2)L) + (kq_(2))/(L)`

=
`(9 * 10^(9))/(0.25)[(-3.3)/(√(2)) + 4.2] * 10^(-6)`

=
`6.72 * 10^(4) V`

Hence, electric potential at point A is
`6.72 * 10^(4) V`.

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

`V_(B) = (kq_(1))/(L) + (kq_(2))/(√(2)L)`

=
`(9 * 10^(9))/(0.25) [-3.3 + (4.2)/(√(2))] * 10^(-6)`

=
`-1.19 * 10^(4) V`

Hence, electric potential at point B is
`-1.19 * 10^(4) V`.