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In: Physics

The drawing shows a positive point charge +q1, a second point charge q2 that may be...

The drawing shows a positive point charge +q1, a second point charge q2 that may be positive or negative, and a spot labeled P, all on the same straight line. The distance d between the two charges is the same as the distance between q1 and the spot P. With q2 present, the magnitude of the net electric field at P is twice what it is when q1 is present alone. Given that q1= +3.87

Solutions

Expert Solution

Given q 1 = 3.87 x 10 -6 C

Electric field at P due to charge q 1 is E = K q 1 / d 2

Where K = 8.99 x 10 9 N m 2/ C 2

Electric field at P due to charge q 2 is E ' = K q 2 / (2d) 2

                                                            = K q 2 / 4(d) 2

                                                            = (1/4)K q 2 / (d) 2

Given net electric field at point P is twice of E

i.e., E + E ' = 2 E

            E ' = 2E - E

           E ' = E

(1/4)K q 2 / (d) 2 = K q 1 / (d) 2

          q 2 / 4 = q 1

                    q 2 = 4 q 1

                           =4 x 3.87

genius_generous answered 7 months ago
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