Question

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

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Solution

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(a)

Charge on sphere A, $q_{A}=6.5×10_{−7}C$

Charge on sphere B, $q_{B}=6.5×10_{−7}C$

Distance between the spheres, $r=50cm=0.5m$

Force of repulsion between the two spheres,

$F=4π∈_{0}r_{2}q_{A}q_{B} $

Where, $∈_{0}=$ Free space permittivity

$4π∈_{0}1 =9×10_{9}Nm_{2}C_{2}$

$∴F=(0.5)_{2}9×10_{9}×(6.5×10_{−7})_{2} $

$=1.52×10_{−2}N$

Therefore, the force between the two spheres is $1.52×10_{−2}N$

(b)

After doubling the charge,

Charge on sphere A, $q_{A}=2×6.5×10_{−7}C=1.3×10_{−6}C$

Charge on sphere B, $q_{B}=2×6.5×10_{−7}C=1.3×10_{−6}C$

Now, if the distance between the sphere is halved, then

$r=20.5 =0.25m$

Force of repulsion between the two sphere,

$F=4π∈_{0}r_{2}q_{A}q_{B} =(0.25)_{2}9×10_{9}×1.3×10_{−6}×1.3×10_{−6} =0.24336N$

Therefore, the force between the two sphere is approximately 0.243 N.

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