Question

Electric field at origin due to a point charge 1 nC placed
at (1, 1)m in x - y plane, is​

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.

Answer 2

Answer:

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} }d=

(y2−y1)

2

+(x2−x1)

2

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

(1−0)

2

+(1−0)

2

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

(1)

2

+(1)

2

\sf \implies d = \sqrt{2} \: m⟹d=

2

m

Now, field intensity be E :

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

d

2

kq

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

9

×

(

2

)

2

10

−9

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

9

×

2

10

−9

\sf \implies\: E =4.5 \: N/C⟹E=4.5N/C

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