Question

Three charges +9uc and -9uc are placed at the corner A of an equilateral triangle of side 3 cm. Find electric field intensity at third vertex.

Answer 1

Given:

Two charges +9uc and -9uc are placed at the corner A of an equilateral triangle of side 3 cm.

To find:

Electric field intensity at third vertex.

Calculation:

Electrostatic Field Intensity due to +9 micro-C:

$\therefore \: | E_{1}| = \dfrac{k \times 9 \times {10}^{ - 6} }{ {(3 \times {10}^{ - 2} )}^{2} }$

$= > \: | E_{1}| = \dfrac{k \times 9 \times {10}^{ - 6} }{ 9 \times {10}^{ - 4} }$

$= > \: | E_{1}| = k \times {10}^{ - 2}$

Electrostatic field intensity due to -9 micro-C:

$\therefore \: | E_{2}| = \dfrac{k \times 9 \times {10}^{ - 6} }{ {(3 \times {10}^{ - 2} )}^{2} }$

$= > \: | E_{2}| = \dfrac{k \times 9 \times {10}^{ - 6} }{ 9 \times {10}^{ - 4} }$

$= > \: | E_{2}| = k \times {10}^{ - 2}$

Hence , we can say that :

$\therefore \: | E_{1}| = |E_{2}| = E \: \: \: ..........(let)$

The net Electrostatic Field Intensity will be :

$\therefore \: E_{net} = \sqrt{ {E_{1}}^{2} + {E_{2}}^{2} + 2E_{1}E_{2} \cos(120 \degree) }$

$= > \: E_{net} = \sqrt{ {E_{1}}^{2} + {E_{2}}^{2} + 2E_{1}E_{2} \times ( - \frac{1}{2} ) }$

$= > \: E_{net} = \sqrt{ {E_{1}}^{2} + {E_{2}}^{2} - E_{1}E_{2} }$

$= > \: E_{net} = \sqrt{ {E}^{2} + {E}^{2} - (E \times E) }$

$= > \: E_{net} = \sqrt{ {E}^{2} }$

$= > \: E_{net} = E$

$= > \: E_{net} = k \times {10}^{ - 2}$

Putting value of Coulomb's Constant:

$= > \: E_{net} = (9 \times {10}^{9} ) \times {10}^{ - 2}$

$= > \: E_{net} = 9 \times {10}^{7} \: N {C}^{ - 1}$

So , final answer is :

$\boxed{ \bold{\: E_{net} = 9 \times {10}^{7} \: N {C}^{ - 1} }}$

Answer 2

Answer:

Explanation:

Each charge will produce the same magnitude of intensity, say E, at the centroid. These are directed at angles of 120 o with each other. So, their vector sum will be zero.