Calculate in the given figure : (i) electric potentials at points A and B, (ii) work done in taking the same
amount of negative charge from A to B, (OA = 0.10 m, OB = 0.20 m).
Answers
(i) electric potentials at points A and B
q = 10μC = 10^-5 C , OA = 0.1 m , OB = 0.2m
electric potential at point A , V_A = Kq/OA
= (9 × 10^9 × 10^-5)/(0.1)
= 9 × 10^5 volts
electric potential at point B, V_B = Kq/OB
= (9 × 10^9 × 10^-5)/(0.2)
= 4.5 × 10^5 volts
(ii) work done in taking the same amount of negative charge from A to B
so change in potential from B to A , ∆V = V_B - V_A = 4.5 × 10^5 - 9 × 10^5 = -4.5 × 10^5 volts
workdone in taking q from A to B = q∆V
= (-10^-5 C)(-4.5 × 10^5)
= 4.5 J
therefore workdone in taking the same amount of negative charge from A to B is 4.5 J
Answer:
i) electric potentials at points A and B
q = 10μC = 10^-5 C , OA = 0.1 m , OB = 0.2m
electric potential at point A , V_A = Kq/OA
= (9 × 10^9 × 10^-5)/(0.1)
= 9 × 10^5 volts
electric potential at point B, V_B = Kq/OB
= (9 × 10^9 × 10^-5)/(0.2)
= 4.5 × 10^5 volts
(ii) work done in taking the same amount of negative charge from A to B
so change in potential from B to A , ∆V = V_B - V_A = 4.5 × 10^5 - 9 × 10^5 = -4.5 × 10^5 volts
workdone in taking q from A to B = q∆V
= (-10^-5 C)(-4.5 × 10^5)
= 4.5 J