Question

Electric field at point P (2 m, 1 m, 2 m) due to a point charge 5.0 μC placed at the origin will be

Answer 1

Given:

Charge of 5.0 μC placed at origin.

To find:

Net electrostatic field intensity at (2 , 1 , 2) metres.

Calculation:

Net distance of that point from origin is:

$d = \sqrt{ {(x2 - x1)}^{2} + {(y - y1)}^{2} + {(z2 - z1)}^{2} }$

$\implies d = \sqrt{ {(2 - 0)}^{2} + {(1 - 0)}^{2} + {(2 - 0)}^{2} }$

$\implies d = \sqrt{4 + 1 + 4}$

$\implies d = \sqrt{9}$

$\implies d = 3 \: metres$

So, net field intensity:

$E = \dfrac{kq}{ {d}^{2} }$

$\implies E = \dfrac{(9 \times {10}^{9}) \times( 5 \times {10}^{ - 6}) }{ {3}^{2} }$

$\implies E = {10}^{9} \times( 5 \times {10}^{ - 6})$

$\implies E =5000 \: N/C$

So, field intensity is 5 × 10³ N/C