Please explain steps (steps of calculation are unclear) ) he electricF(1) 4F(2) 트F(3) Zero-(4)8.444.The electric intensity due to a dipole of length 1010.The electriccm and having a charge of 500 µ, at a point on thetance r frodipole is roaxis at a distance 20 cm from one of the charges inair, is(1) 6.25 × 107 N/C(3) 13.1× 1111 N/Cperpendicpoint will(1) E(2) 9.28 × 107 N/C(4) 20.5 × 107 N/C45.An electrAn electric dipole is situated in an electric field ofE whose dipole moment is p andfield. It e41.(1) A fore(2) A for

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0. The electric intensity due to a dipole of length 10 cm and having a charge of 500 µ, at a point on the axis at a distance 20 cm from one of the charges in air, is (1) 6.25 × 107 N/C (3) 13.1 × 1111 N/c (2) 9.28 × 107 N/C (4) 20.5 × 107 N/C

0. The electric intensity due to a dipole of length 10 cm and having a charge of 500 µ, at a point on the axis at a distance 20 cm from one of the charges in air, is (1) 6.25 × 107 N/C (3) 13.1 × 1111 N/c (2) 9.28 × 107 N/C (4) 20.5 × 107 N/C

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter24: Electric Fields

Section: Chapter Questions

Problem 52PQ

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Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

Please explain steps (steps of calculation are unclear)

) he electric
F
(1) 4F
(2) 트
F
(3) Zero
-
(4)
8.
4
44.
The electric intensity due to a dipole of length 10
10.
The electric
cm and having a charge of 500 µ, at a point on the
tance r fro
dipole is ro
axis at a distance 20 cm from one of the charges in
air, is
(1) 6.25 × 107 N/C
(3) 13.1× 1111 N/C
perpendic
point will
(1) E
(2) 9.28 × 107 N/C
(4) 20.5 × 107 N/C
45.
An electr
An electric dipole is situated in an electric field of
E whose dipole moment is p and
field. It e
41.
(1) A fore
(2) A for

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