How much work is done by an applied force that moves two charges of 7.6 mu C tha

Question

How much work is done by an applied force that moves two charges of 7.6 mu C that are initially very far apart to a distance of 4.6 cm apart? The nucleus of a helium atom contains two protons that are approximately 2.00 fm apart. How much work must be done by an external agent to bring the two protons from an infinite separation to a separation of 2.00 fm? Give your answer in scientific notation. How much work does it take for an external force to set up the arrangement of charged objects in the diagram on the corners of a right triangle when the three objects are initially very far away from each other?

Explanation / Answer

apply Work Done W = Kq1q2/r

where K is constant = 9e9

q1 and q2 are the charges

and r is the distance between them

so

W = 9e9 * 7.6 e -6 * 7.6e -6 /(0.046)

W   = 11.3 Joules

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W = Kq1q2/r

here q1 = q2 = charge of proton = 1.6 e-19 C

so

W = 9e9 * 1.6 e-19 *1.6 e-19 /(2 e-15)

W = 1.152 e-13 J

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net Work Done W = total PE

W = U12 + U23 +U13

so

W = Kq1q2/r1    + kq2q3/r2   + K q1q3/r3

here r1 = 0.12 m

r2 = 0.16 m

r3 = sqrt(0.12^2 +0.16^2) = 0.04 m

so

W =-(9e9 * 5.6 e-6 * 6.1 e -6/0.12) + (9e9 * 6.1 e-6 * 3.6 e-6/0.16) + (9e9 * 5.6 e-6 * 3.6 e-6 /0.04)

W = -3.797 + 4.536

W = 0.739 Joules

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