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Answer To Physic Questions

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Engineering

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Homework

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Question 1:

Answer: f

All of the points A, B, C and D have the same strength of electric field. That is because the

electric fields at A, B, C and D are all 0 due to the definition of uncharged (neutral) spherical

conductor.

Moreover, the electric fields at A, B, C and D can be calculated by using Gauss’s law

( )

enc

A B C D

E d A

E E E E E



=

  =

=

 = = = = =



Question 2:

Answer: f

All of the points E, F, G, H have the same strength of electric field. Moreover, the electric fields

at E, F, G, H can be calculated by using Gauss’s law

( )

enc

total

E F G H

E d A

E E E E E









=

  =

=

 = = = = =



Question 3:

The approximate force on a 5 C charge is calculated by

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( )( )

( )

5 4

kq q



=

Hence, the force is

5 N

Question 4:

a) For

ra

, by applying Gauss’s law, we have

enclosed

E dA





=

  =

=



For

a r b

, by applying Gauss’s law, we have

( )

2 3 3

enclosed

E dA

E A r a

E r r a

















=

  = −



 = −







For

rb

, by applying Gauss’s law, we have

( )

2 3 3

enclosed

E dA

E A b a

E r b a

















=

  = −

−

=



b) For

ra

, the electric potential (V) is zero since the electric field is zero, i.e.,

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Question 1: Answer: f All of the points A, B, C and D have the same strength of electric field. That is because the electric fields at A, B, C and D are all 0 due to the definition of uncharged (neutral) spherical conductor. Moreover, the electric fields at A, B, C and D can be calculated by using Gauss’s law  E d A  qenc  EA 0 0 0 E0  E A  EB  EC  ED   E   0 Question 2: Answer: f All of the points E, F, G, H have the same strength of electric field. Moreover, the electric fields at E, F, G, H can be calculated by using Gauss’s law  E d A  qenc  EA Qtotal 0 0  E  4 R 2  E Qtotal 0 Qtotal 4 R 2 0  EE  EF  EG  EH   E   Question 3: The approximate force on a 5 C charge is calculated by Qtotal 4 R 2 0 kq1q2 r2 1  5 C  4 0 C    2 4 0 1 m  F 5 N Hence, the force is 5 N . Question 4: a) For r  a , by applying Gauss’s law, we have  E  dA  Qenclosed 0  E  4 r  0 E0 2 For a  r  b , by applying Gauss’s law, we have  E  dA  Qenclosed 0 4 3  EA  r  a3  3 0 4 3  E  4 r 2   r  a3  3 0 E   a3  r  2  3 0  r  For r  b , by applying Gauss’s law, we have  E  dA  Qenclosed 0 4 3 3  EA b  a  3 0 4? ...
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