The Electrostatic Force on a Small Sphere of Charge 0.4 μC Due to Another Small Sphere of Charge − 0.8 μC in Air is 0.2 N. (A) What is the Distance Between the Two Spheres? - Physics

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Numerical

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC in air is 0.2 N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

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Solution

(a) Electrostatic force on the first sphere, F = 0.2 N

Charge on this sphere, q1 = 0.4 μC = 0.4 × 10−6 C

Charge on the second sphere, q2 = − 0.8 μC = − 0.8 × 10−6 C

Electrostatic force between the spheres is given by the relation,

`F=(q_1q_2)/(4piin_0r^2)` and , `1/(4piin_0)=9xx10^9 Nm^2C^-2`

Where, ∈0 = Permittivity of free space

`r^2=(q_1q_2)/(4piin_0F)`

`= (9xx10^9xx0.4xx10^-6xx0.8xx10^-6)/0.2`

`=144xx10^-4`

`r=sqrt(144xx10^-4)` = 0.12 m

The distance between the two spheres is 0.12 m.

(b) Both the spheres attract each other with the same force. Therefore, the force on the second sphere due to the first is 0.2 N

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Chapter 1: Electric Charges and Fields - Exercise [Page 46]

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NCERT Physics Class 12
Chapter 1 Electric Charges and Fields
Exercise | Q 1.2 | Page 46
NCERT Physics Class 12
Chapter 1 Electric Charge and Fields
Exercise | Q 1.2 | Page 46

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