Question

Two small spheres A and B of charges 18 esu and 8 esu respectively,
are placed at a distance 20 cm apart in air. At which point, the
electric field intensity will be zero?​

Answer 1

Answer:

I hope I helped you

Explanation:

Initially 2nC and −8nC are separated by a distance d.

Electrostatic force acting between them initially F=

d

2

kq

1

q

2

∴ F=

d

2

(2)(8)k

=

d

2

16k

Now the two charges are allowed to touch each other, they share equal charges among themselves.

Total charge of the system Q=2−8=−6nC

Thus new charge on each sphere q

1

′

=q

2

′

=−3nC

Let the new distance between them be d

′

.

Thus force acting between them F

′

=

(d

′

)

2

kq

1

q

2

′

∴ F

′

=

(d

′

)

2

k(−3)(−3)

=

(d

′

)

2

9k

But F

′

=F

∴

(d

′

)

2

9k

=

d

2

16k

Or (d

′

)

2

=

16

9

d

2

⟹ d

′

=

4

3d