# 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? - Physics

Numerical

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"_"A""q"_"B")/(4piin_0r^2)

Where,

0 = Free space permittivity

1/(4piin_0) = 9 xx 10^9  "N m"^2  "C"^-2

∴ "F" = (9 xx 10^9 xx (6.5 xx 10^-7)^2)/((0.5)^2)

= 1.52 × 10−2 N

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

Concept: Charging by Induction
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 1 Electric Charges and Fields
Exercise | Q 1.12 (a) | Page 46
NCERT Class 12 Physics Textbook
Chapter 1 Electric Charge and Fields
Exercise | Q 12.1 | Page 46
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