A point charge q 2 = -4.2 C is fixed at the origin of a co-ordinate system as sh

Question

A point charge q2 = -4.2 C is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 1 C is is initially located at point P, a distance d1 = 7.7 cm from the origin along the x-axis

3)

What is the potential energy of the system composed of the three charges q1, q3, and q4, when q1 is at point R? Define the potential energy to be zero at infinity.

J

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4)

The charge q4 is now replaced by charge q5 which has the same magnitude, but opposite sign from q4 (i.e., q5 = 2.1 C). What is the new value for the potential energy of the system?

J

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5)

Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-4.2C). Charge q2 is located at the origin and charge q6 is located a distance d = d1 + d2 = 10.8cm from the origin as shown. What is PE, the change in potential energy now if charge q1 is moved from point P to point R?

J

Explanation / Answer

3) PE of two charge system = kq1q2 / d

d is the distance between them.

PE = [ k(q1)(q3) / sqrt(d2^2 + a^2) ] + [k(q1)(q4) / sqrt(d2^2 + a^2) ] + [ k(q3)(q4) / (2a) ]

values are not provided values of q3 and q4 and d2.

so plug in values and get answer.

(put the value of q3, q4 etc, with sign )

4) PE = [ k(q1)(q3) / sqrt(d2^2 + a^2) ] + [k(q1)(q5) / sqrt(d2^2 + a^2) ] + [ k(q3)(q5) / (2a) ]

5)
Pe when q1 is at P:

PEi = [k(q1)(q6) / (d2) ] + [ k(q1)(q2)/(d1) ] + [k(q2)(q6) / (d1 +d2)]

when at R:

PEf = [k(q1)(q6) / (d2 +d1) ] + [ k(q1)(q2)/(d2) ] + [k(q2)(q6) / (d1 +d2)]

deltaPE = PEf - PEi

= [k(q1)(q6) / (d2 +d1) ] + [ k(q1)(q2)/(d2) ] - [k(q1)(q6) / (d2) ] - [ k(q1)(q2)/(d1) ]

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