Part A Find the magnitude of the net electric field these charges produce at poi

Question

Part A

Find the magnitude of the net electric field these charges produce at point A.

Part B

Find the direction of the net electric field these charges produce at point A.

Choose directed to the right or directed to the left.

Part C

Find the magnitude of the net electric field these charges produce at point B

Part D

Find the direction of the net electric field these charges produce at point B.

Choose directed to the right or directed to the left.

Part E

What would be the magnitude of the electric force this combination of charges would produce on a proton at A?

Part F

What would be the direction of the electric force this combination of charges would produce on a proton at A?

Choose directed to the right or directed to the left.

Explanation / Answer

i am allowed to answer only 4 parts at a time

A)
E1 due to -12.5 nC will be towards right (hence positive)
E1 = K*q1/r1^2
= (9*10^9)*(12.5*10^-9) / (0.1)^2 {since r1 = 10 cm = 0.1 m}
= 11250 N/C

E2 due to -6.25 nC will be towards left (hence negative)
E2 = K*q2/r2^2
= -(9*10^9)*(6.25*10^-9) / (0.15)^2 {since r12= 25-10 cm = 15 cm = 0.15 m}
= -2500 N/C

E net = E1 + E2
= 11250 - 2500
= 8750 N/C

Answer: 8750 N/C

B)
Since E net is positive, direction is towards right

C)
E1 due to -12.5 nC will be towards right (hence positive)
E1 = K*q1/r1^2
= (9*10^9)*(12.5*10^-9) / (0.35)^2 {since r1 = 25+ 10 =35 cm = 0.35 m}
= 918.4 N/C

E2 due to -6.25 nC will be towards right (hence positive)
E2 = K*q2/r2^2
= (9*10^9)*(6.25*10^-9) / (0.1)^2 {since r12= 10 cm = 0.1 m}
= 5625 N/C

E net = E1 + E2
= 918.4+ 5625
= 6543.4 N/C

Answer: 6543.4 N/C

D)
Since E net is positive, direction is towards right

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