Question

Two charges +10×10^5 and -10×10^5 are placed 5mm apart determine the electric field

Answer 1

Answer:

ANSWER

Let the two charges be q

1

,q

2

q

1

+q

2

=10

Force between them when placed 2 cm apart

F=

(2×10

−2

)

2

k(q

1

)(q

2

)

F=

(2×10

−2

)

2

k(q

1

)(10−q

1

)

differentiate F and equate it to zero to find at value of q

1

, F will be maximum.

dq

1

dF

=

(2×10

−2

)

2

k

(10−2q

1

)=0

⇒q

1

=5

q

2

=10−q

1

q

2

=5

∴q

1

=q

2

=5 C