Electric field due to a point charge 1 nc placed at (1,1)m in X-Y plane is
1)9N/C
2)4.5N/C
3)9√2N/C
4)9/√2N/C
Answers
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} }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.
Answer: