The figure below shows a test charge q between thetwo positive charges. Find the

Question

The figure below shows a test charge q between thetwo positive charges. Find the force (in newtons) on the testcharge for q = -3 µC.

a)
Give a positive answer if the force is to the right and a negativeanswer if the force is to the left.

2 microCoulombs<---3cm--->q<---2cm--->3microCoulombs b) For the previous question, find the electric field (innewtons/coulomb) at the position of test charge. Again, supply apositive value if the electric field points to the right and anegative value if it points to the left.
SEPERATE QUESTION How much work (in joules) is done in moving a charge of2.5C a distance of 33 cm along an equipotential at12~V?
ANOTHER QUESTION Which of the following are valid units for electric fieldstrength?
The figure below shows a test charge q between thetwo positive charges. Find the force (in newtons) on the testcharge for q = -3 µC.

a)
Give a positive answer if the force is to the right and a negativeanswer if the force is to the left.

2 microCoulombs<---3cm--->q<---2cm--->3microCoulombs b) For the previous question, find the electric field (innewtons/coulomb) at the position of test charge. Again, supply apositive value if the electric field points to the right and anegative value if it points to the left.
SEPERATE QUESTION How much work (in joules) is done in moving a charge of2.5C a distance of 33 cm along an equipotential at12~V?
ANOTHER QUESTION Which of the following are valid units for electric fieldstrength?

Explanation / Answer

(a) .          Electrostaticforce on "q" due to 3 C is : .               F1   = k (3 x 10-6 * - 3 x10-6 ) / ( 2 x 10-2)2 .                      =   ( 9 x 109 * 3 x 10-6 * - 3 x 10-6 ) / ( 2 x 10-2 )2 .                       =  - 202.5   N .          Electrostaticforce on "q" due to 2 C is : .               F1   = - k (2 x 10-6 * - 3 x10-6 ) / ( 3 x 10-2)2 .                      =   ( 9 x 109 *  2x 10-6 * 3 x 10-6 ) /( 3 x 10-2 )2 .                       =  60    N . Hence net force on "q" is : .              F     = F1   + F2   =   - 202.5   N  +   60    N = - 142.5 N . (b) .         Electric field is: .             E1    =   ( 9 x109 * 3 x 10-6  ) / ( 2 x 10-2 )2 .                     = 6.75 x 107    N/C .              E2   =  - ( 9 x 109 *  2x 10-6   ) / ( 3 x 10-2 )2 .                     =  - 2 x 107    N/C .         Net field is: .            E       =  E1    + E2   = 6.75 x 107    N /C   - 2 x 107    N/C    =  4.75 x 107    N /C .          (c) .          We know that: .               V   = W / Q .          or W   = V * Q   = 0    (since it is equipotential surface ) .                                         =   0  J . Hope this helps u! .               F1   = - k (2 x 10-6 * - 3 x10-6 ) / ( 3 x 10-2)2 .                      =   ( 9 x 109 *  2x 10-6 * 3 x 10-6 ) /( 3 x 10-2 )2 .                       =  60    N . Hence net force on "q" is : .              F     = F1   + F2   =   - 202.5   N  +   60    N = - 142.5 N . (b) .         Electric field is: .             E1    =   ( 9 x109 * 3 x 10-6  ) / ( 2 x 10-2 )2 .                     = 6.75 x 107    N/C .              E2   =  - ( 9 x 109 *  2x 10-6   ) / ( 3 x 10-2 )2 .                     =  - 2 x 107    N/C . .                     =  - 2 x 107    N/C .         Net field is: .            E       =  E1    + E2   = 6.75 x 107    N /C   - 2 x 107    N/C    =  4.75 x 107    N /C .          (c) .          We know that: .               V   = W / Q .          or W   = V * Q   = 0    (since it is equipotential surface ) .                                         =   0  J . Hope this helps u!

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