Question

Two point charges, each having a charge 6.5×10-⁷ coulomb is separated by a distance of 75 cm in vacuum. Calculate the force of repulsion between them.​

Answer 1

Answer:

Charge on the first sphere, q

1

=2×10

−7

C

Charge on the second sphere, q

2

=3×10

−7

C

Distance between the spheres, r=30cm=0.3m

F=

4πε

0

r

2

q

1

q

2

Where, ε

0

= Permittivity of free space

4πε

0

1

=9×10

9

Nm

2

C

−2

F=

(0.3)

2

9×10

9

×2×10

−7

×3×10

−7

=6×10

−3

N

Hence, force between the two small charged spheres is 6×10

−3

N. The charges are of same nature. Hence, force between them will be repulsive.

Answer 2

• Explanation:

F=kQ*Q/r²

F=k6.5 * 10 की पावर माइनस 7 * सिक्स पॉइंट 5 * 10 की पावर माइनस 7 डिवाइडेड बाय 0.75²

F =9*6.5*6.5*10ka power minus 6 /0.75*0.75

9*65*65*10 ka pawer minus 4/75*75

F=6.7*10-⁴