Two positive point charges are initially separated by adistance of 2 cm. If thei

Question

Two positive point charges are initially separated by adistance of 2 cm. If their separation is increased to 6 cm, theresultant electrical potential energy is equal to what factor timesthe initial electrical potential energy? Two positive point charges are initially separated by adistance of 2 cm. If their separation is increased to 6 cm, theresultant electrical potential energy is equal to what factor timesthe initial electrical potential energy? Two positive point charges are initially separated by adistance of 2 cm. If their separation is increased to 6 cm, theresultant electrical potential energy is equal to what factor timesthe initial electrical potential energy?

Explanation / Answer

     The electric potential energy is U = k*q /r2                                        U 1 / r2       U1 / U2 = (r2 /r1)2                                            = ( 6 cm /2cm)2                                          = 9

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