Two charges are placed as shown in figure 1 with q126 uC and

Two charges are placed as shown in figure 1 with q1=2.6 uC and q2=-5.3uC. Find the potential difference between points A and B.

The electric potential due to charge q1 at point A V_1a

V_1a = k q1 / x_1a ; x1a is the distance of q1 from A = 0.10m

= 8.98*10^9 * 2.6 *10^-6 / 0.1 m

= 2.3348*10^5 V

simillarly

The electric potential due to charge q2 at point A V_2a

V_2a = k q2 / x_2a ; x_2a is the distance of q2 from A

we can determine x_2a by phythogorus law from the triangle formed from q2 to A and to B

x_2a =sqrt (0.1^2+0.1^2) = 0.141 m

V_2a = k q2 / x2a

= 8.98*10^9 * -5.3 *10^-6 / 0.141 m

= -4.75940* 10 ^5 volt

net potential due to q1 and q2 at point A VA = V_1a + V_2a =2.3348*10^5 V -4.75940* 10 ^5 volt

VA = -2.424 * 10^5 Volt

The electric potential due to charge q2 at point B V_2b

V_2b = k q2 / x_2b ; x2b is the distance of q2 from B = 0.1 m

= 8.98*10^9 * (-5.3) *10^-6 / 0.1 m

= -3.37546 * 10^ 5 V

The electric potential due to charge q1 at point B V_1b

V_1b = k q1 / x_2b ; x2b is the distance of q2 from B = 0.141 m which can be calculated same as the above formulation

= 8.98*10^9 * (2.6) *10^-6 / 0.141 m

= 1.65588 * 10^ 5 V

net potential at point B VB = V_1b+V_2b = 1.65588 * 10^ 5 V -3.37546 * 10^ 5 V = -1.719 * 10^5 volt

net electric potential difference between A and B = VA -VB = -2.424 * 10^5 +1.719 * 10^5

= - 7.09*10^4