The potential difference between two points separated by 50 cm in the field of source charge 500 ×10^-6 is 400 V. The electric field intensity of the charge isa. 20000 N/Vb. 400 V/mc. 2000 V/m(correct answer is (a). kindly give proper reason with solution otherwise don't answer)
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Answer:
REF.Image
Consider at point P, the electric field is zero,
as Electric field,
E=
d
2
kq
at P,
x
2
kq
1
=
(50−x)
2
kq
2
⇒
x
2
4
=
(50−x)
2
9
⇒(
x
50−x
)
2
=
4
9
⇒
x
50−x
=±
2
3
⇒x=20cm OR x=−150cm (Not Applicable)
Hence, at a distance = 20 cm from A, E=0
Also, potential at that point,
V=
x
kq
1
+
50−x
kq
2
=9×10
9
×(
0.2
4×10
−6
+
0.3
9×10
−6
)=4.5×10
5
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