Two spherical point charges each carrying a charge of 40 mC are attached to the

Question

Two spherical point charges each carrying a charge of 40 mC are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1, what is the length of the spring when the charges are in equilibrium?

Two points A and B lie 20 cm apart on a line extending radially from a point charge Q. If the potential at A is 150 V and the potential at B is 120 V, find the charge Q and the distance of point A from the charge. What is the work that needs to be done to move a 10 µC charge from B to A?

Explanation / Answer

Here,

let the length of spring when charges are in equilibrium is x

charge , q = 40 mC

spring constant , k = 120 N/m

for equilibrium

k * q^2/x^2 = k * (x - 0.20)

9*10^9 * (0.040)^2/x^2 = 120 * (x - 0.20)

solving for x

x = 49.4 m

the final length of spring needed is 49.4 m

--------------------------------------------------------

let the charge is q

distance of point A is r

Now , potential at A = 9*10^9 * Q/r

9*10^9 * Q/r = 150

potential at B

9*10^9 * Q/(r + 0.20) = 120

solving for Q and r

Q = 1.33 *10^-8 C

r = 0.8 m

the point charge is 1.33 *10^-8 C

point A is 0.8 m away from the charge

work needed to move the charge = 10 *10^-6 * (150 - 120)

work needed to move the charge = 3 *10^-4 J

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