Electric field intensity at a point B due to point charge Q kept at a point A is 24N/C and the electric potential at the point B due to the same charge is 12J/C. Calculate the distance AB and the magnitude of charge Q.
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Answered by
73
E = K Q / AB^2 = 24 N/C
V = K Q / AB = 12 J/C
So AB = V/E = 12/24 = 0.50 meters
So V = 9*10^9 * Q / 0.50 = 12 J/C
Q = 2.67 nano Coul.
V = K Q / AB = 12 J/C
So AB = V/E = 12/24 = 0.50 meters
So V = 9*10^9 * Q / 0.50 = 12 J/C
Q = 2.67 nano Coul.
Answered by
17
Answer:
Distance=0.5m
Charge=6.67*10^-10
Explanation:
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