A charge Q is located at a grid intersection in the x-y plane (see plot below),

Question

A charge Q is located at a grid intersection in the x-y plane (see plot below), with Q = -2.30 mu C. The electric potential in the x-y plane due to a charge distribution near the origin is given by V(x, y) = beta x/(x^2 + y^2)^3/2, with beta = 4.808 times 10^5 Nm^3/C, x and y in m. Calculate the x-component of the electric force on Q. II. The electric potential v(x, y) is calculated at the locations N and U using the equation above. Its values are V_N = 4.480 times 103^volts and v_U = 4.116 times 1^03 volts. Calculate the y component of the force on Q.

Explanation / Answer

a. Consider all differentiationa as partial differentiations.

E   =   - dV/dr

            =   - ( i^ d/dx +    j^ d/dy) { * x / (x2 +   y2)3/2}

            =   * [ i^ (y2 - 2 * x2)   +   j^ (x2 - 2 * y2)] / (x2 + y2)5/2

hence component of E over y axis   Ey   =      * {(x2 - 2 * y2) / (x2 + y2)5/2}

   Substituting      =   4.808 * 105   N/C,   x   = 7 m,   y   = 6 m

   Ey   =     4.808 * 105 * { (72 - 2 * 62) / (72 + 62)5/2

         = -166.02 N/C

   Hence         Fy   =   q * Ey

                              = -2.3 * 10-6 * (-166.02)

                              = 0.382 mN

   b.   Over x axis,   Ex   =    (VU -   VN) / xNU

                                        =   (4.480*103   -   4.110 * 103) / (6.2 - 5.8)

                                       = 910 N/C

   x component of force   Fx   =   q * Ex

                                                =   -2.3* 10-6 * ( 910 * 103)

                                                =   - 2.093m N

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