Two point changes are separated by 25.0 cm (Figure 1) Assume that gt = 8.00 n C

Question

Two point changes are separated by 25.0 cm (Figure 1) Assume that gt = 8.00 n C and = 10.0 nC Find the net electric these charges produces at point. Find the net electric field these charges produce at point B. What would be the magnitude of the electric force this combination of charges would produce on a proton at A? What would be the direction of the electric force this combination of changes would produce on a proton at A? The force would be direction to the left. The force would be directed to the right.

Explanation / Answer

step;1

Given that

Charge q1=-8nC

charge q2=-10nC

step;2

now we find the net electricfield at point A

the electric field from A to charge q1=>E1=9*10^9*-8*10^-9/(0.15)^2=-3200 N/c

the electric field from A to Charge q2=>E2=9*10^9*-10*10^-9/(0.1)^2=-9000 N/c

the net electric field at point A=> Enet=E1+E2=-3200-9000=-12200 N/c

step;3

now we find the electric field at point B

the electric field at point B to charge q1=>E1=9*10^9*-8*10^-9/(0.1)^2=-7200 N/c

the electric field at point B to charge q2=>E2=9*10^9*-10*10^-9/(0.35)^2=-734.7 N/c

the net electric field at point B=>Enet=-7200+734.7=-7934.7 N/c

step;4

now we find the net force at point Ais proton of charge Q

the force of attraction b/w Q1 and Q

force F1=9*10^9*1.6*-8*10^-9*10^-19/(0.15)^2

=-5120*10^-19 N

force F2=9*10^9*1.6*-10*10^-19*10^-9/(0.1)^2=-14400*10^-19 N

the net force of attraction at point B=>Fnet=[-5120-14400]*10^-19=-19520*10^-19 N

step;5

the force attraction is left side

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