Question

Please help me to solve this problem??.....???? ​

Answer 1

Answer:

Net electric field at A due to charges at B and C is

E

A

=2E

Ac

sin50

o

=2×9×10

9

×

(0.20)

2

3×10

−6

×

2

3

=

3

×6.75×10

5

AM=

(20)

2

−(10)

2

=

400−100

=10

3

cm

Let the charge at M be q. Charge q should be positive so that there can be repulsion between the charges at A and M.

E

AM

=9×10

9

×

(10

3

/100)

2

q

=3q×10

11

For A to be in equilibrium

E

A

=E

AM

⇒

3

×6.75×10

5

=3q×10

11

or q=

4

9

3

×10

−6

=

4

9

3

μC