Question

two charges +25 NC & -25 NC are placed 6m apart find the intensity of electric field​

Answer 1

Given:

Two charges +25 nC and -25 nC are placed 6m apart.

To find:

Net field Intensity at the midpoint of the line joining the charges.

Calculation:

Distance of each charge from midpoint is 3m.

So, net field intensity:

$E = \dfrac{k \times 25 \times {10}^{ - 9} }{ {3}^{2} } \: \hat{i} + \dfrac{k \times 25 \times {10}^{ - 9} }{ {3}^{2} } \: \hat{i}$

$= > E = 2 \bigg \{\dfrac{k \times 25 \times {10}^{ - 9} }{ {3}^{2} } \bigg \} \: \hat{i}$

$= > E = 2 \bigg \{\dfrac{k \times 25 \times {10}^{ - 9} }{ 9} \bigg \} \: \hat{i}$

$= > E = \bigg \{\dfrac{k \times 50 \times {10}^{ - 9} }{ 9} \bigg \} \: \hat{i}$

Putting value of Coulomb's Constant:

$= > E = \dfrac{9 \times {10}^{9} \times 50 \times {10}^{ - 9} }{ 9} \: \hat{i}$

$= > E = \dfrac{ \cancel9 \times {10}^{9} \times 50 \times {10}^{ - 9} }{ \cancel9} \: \hat{i}$

$= > E = 50 \: \hat{i} \: \: N{C}^{ - 1}$

So, final answer is :

$\boxed{ \bold{ \red{ E = 50 \: \: N{C}^{ - 1} }}}$