Question

Two point charges 4*10^-6C and 2*10^-6C are placed at the vertices A and B of a right angled triangle ABC respectively. B is the right angle, AC=2*10^-2 m and BC=10^-2m. Find the magnitude and direction of resultant electric intensity at C?​

Answer 1

Answer:

Let 4C charge which is placed at C has an electric field intensity $$-E_{1}\hat{i} at position B

and electric field intensity due to 9C charge which is placed at position A is $$-E_{2}\hat{j} at position B

⇒E

1

=

r

2

Kq

1

=

2

2

×10

−4

9×10

9

×4×10

−9

=9×10

4

N/C.

⇒E

2

=

r

2

Kq

2

=

3

2

×10

−4

9×10

9

×9×10

−9

=9×10

4

N/C.

resultant E=−E

1

i

^

−E

2

j

^

=−9×10

4

i

^

−9×10

4

j

^

=

9

2

+9

2

×10

4

.

⇒3

2

×10

4

N/C