Question

5. The positive charges InC and 4nC are placed at the points A and C respectively of a right angled
triangle ABC in which B=90° AB=3cm and BC=2cm. Find the magnitude and direction of the
resultant electric intensity at B.
FITCTPIC POTENTIAL AND CAPACITANCE​

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