Question

two point charges each of 20 microcoulomb are placed 50 cm apart in air what is the electric field intensity at the mid point on the line joining the centre of the two point charges

Answer 1

Given:

Two 20 uC charges are seperated by 50 cm.

To find:

Electric field at midpoint of two charges ?

Calculation:

• The midpoint is seperated from each charge by 50/2 = 25 cm = 0.25 m from each charge.

• Now, the field intensity of once charge will be directed opposite to the other charge.

So, net field intensity is:

$E_{net} = \dfrac{k q_{1} }{ {d}^{2} } - \dfrac{k q_{2} }{ {d}^{2} }$

• Here , d = 0.25 metres and q1 = q2 = 20 × 10^(-6) C.

$\implies E_{net} = \dfrac{k \times 20 \times {10}^{ - 6} }{ {(0.25)}^{2} } - \dfrac{k \times 20 \times {10}^{ - 6} }{ {(0.25)}^{2} }$

$\implies E_{net} = 0$

So, net field intensity at midpoint will be zero.

Answer 2

Answer:

Explanation:

= 9×10*9×20×10*-6÷(50×50×10*-4)