I just need Part D. Long Answer (15 14. Use the following diagram to answer all

Question

I just need Part D.  

Long Answer (15 14. Use the following diagram to answer all parts. a. If the ' point charge is really just neutral, what Electric field is created at'X'5.00 C from the other two charges? 25.0 cm 15.0 c1n 5.00 pC b. Ag an if the'?' point charge is zero, what is the electric potential at ? If at'X' a-2.50 C, 10.5 ng test charge were released from rest how fast would it be going wr it crossed the line directly between the two 5.00 C charges? C. d. Finally what charge would point charge ? need such that the test charge wouldn't move at all? 3

Explanation / Answer

(d) aNSWER To part a is 9.08 x 10^5 N/C

that is field due to these two charge.

so we will need third charge, due to which field strength will be 9.08 x 10^5 N/C and to the left/

E = k q / r^2

9.08 x 10^5 = (9 x 10^9) (q)/ 0.25^2

q = 6.31 x 10^-6 C.

Ans: q = - 6.31 uC

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