Two charges of q1 = 1.2 mu C and q_2 = -2.4 mu C are d = 0.64m apart at two vert

Question

Two charges of q1 = 1.2 mu C and q_2 = -2.4 mu C are d = 0.64m apart at two vertices of an equilateral triangle as in the figure below. (a) What is the electric potential due to the 1.2 mu C charge at the third vertex, point p? (b) What is the electric potential due to the -2.4 mu C charge at p? (c) Find the total electric p? (d) What is the work required to move a 3.8 mu C charge from infinity to p? An electron is fired at a speed v_0 = 4.60 times 10^6 m/s and at an angle theta_0 = -45 degree between two parallel conducting plates that are D = 2.5mm apart, as the voltage difference between the plates is Delta nu = 115 V.

Explanation / Answer

electric potential is given

P = kQ/r

total potential = 9*10^9*(1.2*10^(-6)/0.64 - 2.4*10^(-6)/0.64) = - 16875 volt

work done = total potential*charge = 16875*3.8*10^(-6) = 0.0641 J

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