?two identical spheres having charges of opposite sign are placed 3m a...
Ans.
Method to Solve :
Here,
F1 = 0.108 N
r = 0.5 m
F2 = 0.036 N
Let charges on the spheres be q1 and q2 respectively.
Charges after connection gets redistributed in the spheres, so be q in each of the spheres
From conservation of charge
q1 + q2 = 2q ---1.
Again
F1 = k q1q2/r2 ---2
F2 = k q2/r2 ----3
=> 0.036 = 9�109�q2/(0.5)2
=> q2 = 0.036�(0.5)2/9�109
=> q2 = 0.036/(4�9�109)
=> q2 = 1�10-12
=> q = 1�10-6 C ---4
From 2 and 3
F1/F2 = q1q2/q2
=> 3 = q1q2/q2
=> 3q2 = q1q2
=> q1q2 = 3q2
Again,
(q1 – q2)2 = (q1 + q2)2 – 4q1q2
= (2q)2 – 3q2/ r2
= q2/ r2
By 4
(q1 – q2) = q/r
= 1�10-6/ 0.5
= 2�10-6 ---5
From 1 and 4.
q1 + q2 = 4�10-6 ---6.
Adding 5 and 6
q1 = 3�10-6
= 3μC
Subtracting 5 from 6
q2 = 1�10-6
= 1μC
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?two identical spheres having charges of opposite sign are placed 3m a...
Initial Charges on Spheres
- Let's denote the charges on the spheres as q1 and q2, where q1 is positive and q2 is negative.
Initial Force between Spheres
- The force between the spheres can be calculated using Coulomb's law: F = k * |q1 * q2| / r^2, where k is the electrostatic constant, r is the separation distance.
- Given F = 2.4 * 10^-3 N, r = 3m, k = 9 * 10^9 Nm^2/C^2.
- Substituting the values, we get 2.4 * 10^-3 = (9 * 10^9) * |q1 * q2| / 3^2.
Charging by Induction
- When the spheres are touched, they will equalize their charges due to the principle of charging by induction.
- The negative sphere will transfer some charge to the positive sphere until they both have the same charge magnitude.
Final Force between Spheres
- After touching and separating the spheres, the force between them becomes 10^-3 N.
- Using the same formula as before, 10^-3 = (9 * 10^9) * |q1 * q2| / 3^2.
- Since the spheres now have the same charge, q1 = -q2.
- Solving the two equations simultaneously, we can find the initial charges on the spheres.
Therefore, by solving the equations, we can determine the initial charges on the spheres.
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