Question

Electric field at origin due to a point charge 1 nC placed at (1, 1)m in x - y plane, is
9 N/C
92N/C
ENC
N/C
4.5 N/C​

Answer 1

Given:

Point charge of 1 nC is placed at (1,1) point.

To find:

Electrostatic Field Intensity at origin?

Calculation:

Separation between the points:

$\sf d = \sqrt{ {(y2 - y1)}^{2} + {(x2 - x1)}^{2} }$

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

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

$\sf \implies d = \sqrt{2} \: m$

Now, field intensity be E :

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

$\sf \implies\: E =9 \times {10}^{9} \times \dfrac{ {10}^{ - 9} }{ {( \sqrt{2}) }^{2} }$

$\sf \implies\: E =9 \times {10}^{9} \times \dfrac{ {10}^{ - 9} }{2 }$

$\sf \implies\: E =4.5 \: N/C$

So, field intensity is 4.5 N/C at origin.