First: Q1 = -2C to (0,0) Second: Q2 = + 3C to (5,0) a) How much work was done? W

Question

First: Q1 = -2C to (0,0) Second: Q2 = + 3C to (5,0)

a) How much work was done? Was this work done by or against the electric force? (Write answer in exponential notation, e.g. E-4)

At point "p": (3m from Q1; 4m from Q2)

b) What is the potential, V after these charges are moved to these positions?

c) What is the electric field, E at point "p"?

First: Q1 = -2C to (0,0) Second: Q2 = + 3C to (5,0) a) How much work was done? Was this work done by or against the electric force? (Write answer in exponential notation, e.g. E-4) At point p: (3m from Q1; 4m from Q2) b) What is the potential, V after these charges are moved to these positions? c) What is the electric field, E at point p? Two charges are moved from far away ( r space equal space infinity) into their positions as follows

Explanation / Answer

work done in bringing Q1 at (0,0) = 0 Jule

Reason: no electric field is there

work done in bringing Q2 at (5,0) = k*(Q1)*(Q2) / r = - 1.08 * 10^(9) Jules

total work done = 0 + (- 1.08 * 10^(9)) = - 1.08 * 10^(9) Jules

this work is done by the electric force because the force between the charges is attractive in nature.

(b):

potential at point P =  k*(Q1) / r' +  k(Q2) / r'' = - 1.5 * 10 ^(9) V

(c):

electric field at point P = ( k*(Q1) / r' ^2 )^2 + ( k(Q2) / r'' ^2 )^2= 2.6168 * 10 ^(9) V/m

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