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Please respond with detailed steps so that I can understand better, and be able to do similar problems. THANKS.

Consider charged particles Q1= 70.0 uC, Q2= 48.0 uC , Q3= -80.0 uC  which are placed in a straight line along the x-axis. Q2 is placed in the center at a distance of 0.350 m from each of the others, and Q3 is also 0.350 m from some point P, as shown in the figure. (There is no charge located at P).

a) Calculate the net force F1 exerted on Q1

b) Calculate the electric field Ep at point p due to the three charges

c) Calculate de potential V at point P due to the three charges

Explanation / Answer

here ,

Q1= 70.0 uC

Q1 = 70 * 10^-6 C

Q2= 48.0 uC

Q2 = 48 * 10^-6 C

Q3= -80.0 uC

Q3 = - 80 * 10^-6 C

(a)

F = K * Q1 * Q2 / r^2

the net force F1 = K * Q1 * Q2/0.35^2 + k * Q1 * Q2 /0.7 ^2

F1 = (9* 10^9* 70 * 10^-6 * 48 * 10^-6) / 0.35^2 - (9* 10^9* 70 * 10^-6 * 80* 10^-6) / 0.7^2

F1 = 144 N

the net force F1 exerted on Q1 is 144 N in -ve X direction

(b)

E = K * Q/r^2

the net electric feild at P , Ep = K * Q1 / 1.05^2 + K * Q2 / 0.7^2 + K * Q3 / 0.35^2

E = 9 * 10^9 * 10 ^-6 * (70/1.05^2 + 48/0.7^2 - 80/0.35^2)

E = - 4.42 * 10^6 N/C

the electric field Ep at point p due to the three charges is 4.42 * 10^6 N/C in negative X direction

(c)

V = K* Q/r

potential V at point P due to the three charges , V = K * Q1 / 1.05 + K * Q2 / 0.7 + K * Q3 / 0.35

V = 9 * 10^9 * 10 ^-6 * (70/1.05+ 48/0.7 - 80/0.35)

V = - 8.4 * 10^5 V

potential V at point P due to the three charges is - 8.4 * 10^5 V

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