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Two Insulated Charged Copper Spheres a and B Have Their Centers Separated by a Distance of 50 Cm. What is the Mutual Force of Electrostatic Repulsion If the Charge on Each is 6.5 × 10^−7 C? the Radii of a and B Are Negligible Compared to the Distance of Separation - CBSE (Science) Class 12 - Physics

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Question

Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation

Solution

Charge on sphere A, qA = Charge on sphere B, qB = 6.5 × 10−7 C

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

Force of repulsion between the two spheres,

`F=(q_Aq_B)/(4piin_0r^2)`

Where,

0 = Free space permittivity

`1/(4piin_0)=9xx10^9 Nm^2C^-2`

`F=(9xx10^9xx(6.5xx10^-7)^2)/((0.5)^2)`

= 1.52 × 10−2 N

Therefore, the force between the two spheres is 1.52 × 10−2 N.

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APPEARS IN

 NCERT Solution for Physics Textbook for Class 12 (2018 to Current)
Chapter 1: Electric Charge and Fields
Q: 12.1 | Page no. 46
Solution Two Insulated Charged Copper Spheres a and B Have Their Centers Separated by a Distance of 50 Cm. What is the Mutual Force of Electrostatic Repulsion If the Charge on Each is 6.5 × 10^−7 C? the Radii of a and B Are Negligible Compared to the Distance of Separation Concept: Charging by Induction.
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