See attached jpg file for question.
See attached file.
q1 =1.1 μC
q2 = -2 μC
q3 = 1.1 μC
Distance between two charges q1 and q3 can be calculated as
q1 q3 distance = (0.25 + 0.25)1/2
= 0.707 m
So distance between q1 and A will be r1 = 0.707/2
r1 = 0.353 m
Distance between charge q2 and A will also be same as above (r2) = 0.353 m
Distance between charge q3 and A will be r3 = 0.353 m
Potential at point A due to Charge q1 = K*(q1/r1)
Value of K is equal to 1/40
where o is permittivity of free space .Its volume is 8.85 x 10-12 c2/Nm2.
Thus in S.I. system numerical value of K is 8.98755 x 109 Nm2c-2.
So V1 = 8.98755 * 109 *(1.1*10-6)/ (0.353)
= 28.006*103 Volt
Potential at point A due to Charge q2, V2= K*(q2/r2)
= 8.98755 * 109 *(-2*10-6)/ (0.353)
V2 = - 50.92*103 Volt
Potential at point A due to Charge q3, V3 = K*(q3/r3)
V3 = 8.98755 * 109 ...
The expert calculates the net electric potentials due to three charges. A ball with a charge being placed in a field is determined. With careful explanations and complete formulas and calculations, the problem is solved.