Solid Conducting Spheres

The battery can be thought of as a pump of charge that transfers a certain amount of charge from sphere 2 to sphere 1. Let this amount be $Q$ . A negative charge ( $-Q$ ) is induced in sphere 2 while a positive charge( $Q$ ) is induced in sphere 1. Moreover, Since the spheres are connected by a battery, the potential difference between sphere 1 and sphere 2 must be $V$ .

$\implies \frac{Q}{4\pi\epsilon_oR_1} = -\frac{Q}{4\pi\epsilon_oR_2}+V$

This gives the expression for $Q$ in terms of the other variables. Finally, the magnitude of the force between the spheres is:

$F = \frac{Q^2}{4\pi\epsilon_od^2}$

Evaluating all expressions gives the required answer: $\boxed{10^4F=34.906}$ .

An important thing to note is that due to conservation of charge, the charge leaving one sphere will be equal to the charge accumulating on the other. The surface potentials are then:

$V_1 = \frac{k q}{R_1} \\ V_2 = -\frac{k q}{R_2}$

The difference between the surface potentials is equal to the battery voltage:

$V_1 - V_2 = \frac{k q}{R_1} + \frac{k q}{R_2} = V$

Solve for $q$ . Then the force between the spheres is:

$F = \frac{k q^2}{d} \approx \frac{34.9}{10000}$

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import math

R1 = 2.0
R2 = 1.0
V = 10.0
d = 400.0
e0 = 1.0

k = 1.0/(4.0*math.pi*e0)

left = k/R1 + k/R2

q = V/left

F = k*q*q/(d**2.0)

print F
print (10000.0*F)

#>>> 
#0.00349065850399
#34.9065850399
#>>>