A particle with a charge of 5.5 × 10 –8 C isfixed at the origin. A particle with

Question

A particle with a charge of 5.5 × 10–8C isfixed at the origin. A particle with a charge of–2.3 ×10–8C is moved from x = 3.5 cm on thex axis to y = 4.3 cm on the y axis. Thechange in potential energy of the two-particle system is:

3.1 × 10–3 J

–3.1 × 10–3 J

6.0 × 10–5 J

–6.0 × 10–5 J

0

A particle with a charge of 5.5 × 10–8C isfixed at the origin. A particle with a charge of–2.3 ×10–8C is moved from x = 3.5 cm on thex axis to y = 4.3 cm on the y axis. Thechange in potential energy of the two-particle system is:

3.1 × 10–3 J

–3.1 × 10–3 J

6.0 × 10–5 J

–6.0 × 10–5 J

0

Explanation / Answer

Charge q = 5.5 * 10 ^ -8 C

Potential at x = 3.5 cm ( =0.035 m ) is V = Kq / x

                                                                   =( 8.99 * 10 ^ 9 ) ( 5.5*10^-8) / 0.035

                                                                   =14127.14 Volt

Potential at y=4.3 cm ( =0.043 m ) is V ‘ = Kq /y

                                                                   =( 8.99 * 10 ^ 9 ) ( 5.5*10^-8) / 0.043

                                                                   =11498.8 Volt

Potential difference dV = V ~ V ‘

                                      = 2628.3 Volt

Work done W = dV * q ‘

Where q ‘ = -2.3 * 10 ^ -8 C

Plug the values we get W = -6.045 * 10 ^ -5 J

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