Find the force on the charge +qin Fig. 3.14. (The xyplane is a grounded conductor.)

Short Answer

Expert verified

Answer

The net force on q is -14πε0(29272d2)z.

Step by step solution

01

Given data

Consider the given figure as shown below.

Here, from the given figure it is clear that xyplane is grounded conductor, thus potential is zero

02

Determine force

As +qinduces an equal and opposite charge of -qat a distance of Z=-3dand -2qinduces +2qat distance of Z=-d.

Write the expression for force on +qdue to -2q,.

F1=14πε0q(-2q)(2d)2z …… (1)

Write the expression for force on +qdue to +2q,.

F1=14πε0q(-2q)(4d)2z …… (2)

Write the expression for force on +qdue to -q.

F1=14πε0q(-2q)(6d)2z …… (3)

03

Determine the net force

Write the expression for the net force on q .

F=F1+F2+F3=q4πε0[-2q2d2+2q4d2+-q6d2]z=q24πε0[-12+18-136]z=q24πε0(29q272d2)z

Thus, the net force on q is -q24πε0(29q272d2)z.

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