Question

Charges of 2.0 x 10-6 C and 1.0 x 106 C are placed at the corners A and B of square of side 5.0
cm as shown in figure. How much work will be done in moving a charge of 1.0 x 10 C from
to D against the electric field?
​

Answer 1

V = q/C = qK/r = (9×10^9q/r)

Vc = 9×10^9×10^-12×100/5 + 9×10^9×10^-12×2×100/root(50)

Vc = 0.18 (1+root(2))

Similarly Vd = 0.18 × (2+1/root(2))

W = Vd - Vc

W = 0.18(1+root(2)-2-1/root(2))

W = 0.053 J.

Answer 2

Answer:

Work done in moving a charge of $1.0 \times 10^{-6} C$ from C to D is 0.053J

Explanation:

[Assuming all the given charges are actually in microns, i.e., $10^{-6}$C]

Given the charges, $q_{A} =2.0 \times 10^{ -6} C$ and $q_{B} =1.0 \times 10^{ -6} C$        

       side of a square,  $r=5.0cm=5\times 10^{-2} m$                                              

Charge to be moved from C is $1.0 \times 10^{-6} C$

Since the charges are at A and B of a square, the potential at C and D will be the sum of potentials at A and B.

Potential at C is, due to the potential of charge A at diagonal and charge B at distance $r$.

Diagonal distance of a square is $\sqrt{r^{2} +r^{2}}=\sqrt{5^{2} +5^{2}}=\sqrt{50} =5\sqrt{2}$

                          $\therefore \ V_{C} = \frac{kq_{A} }{r_{AC}}+ \frac{kq_{B} }{r_{BC} }= k \Big[ \frac{q_{A} }{r_{AC}}+ \frac{q_{B} }{r_{BC} } \Big]$

                                    $= 9\times 10^{9} \Big[ \frac{2\times 10^{-6} }{5\sqrt{2} \times 10^{-2} }+ \frac{1\times 10^{-6} }{5 \times 10^{-2} } \Big]$

                                   $= 9\times 10^{9} \Big[ \frac{2\times 10^{-4} }{5\sqrt{2} }+ \frac{1\times 10^{-4} }{5} \Big]$

Similarly potential at D is, due to the potential of charge B at diagonal and charge A at distance $r$.

                          $\therefore \ V_{D} = k \Big[ \frac{q_{A} }{r_{AD}}+ \frac{q_{B} }{r_{BD} } \Big]$

                                    $= 9\times 10^{9} \Big[ \frac{2\times 10^{-6}}{5\times 10^{-2} }+ \frac{1\times 10^{-6} }{5\sqrt{2}\times 10^{-2}} \Big]$

                                    $= 9\times 10^{9} \Big[ \frac{2\times 10^{-4} }{5 }+ \frac{1\times 10^{-4} }{5\sqrt{2}} \Big]$

Work done in moving a charge is given by

                              $W=q\Delta V=q(V_{D}-V_{C} )$

             $=1.0 \times 10^{-6}\Big[ 9\times 10^{9} \Big( \frac{2\times 10^{-4} }{5 }+ \frac{1\times 10^{-4} }{5\sqrt{2} } }\Big) - 9\times 10^{9} \Big( \frac{2\times 10^{-4} }{5\sqrt{2} }+ \frac{1\times 10^{-4} }{5} } } \Big) \Big]$

             $=1.0 \times 10^{-6}\times 9\times 10^{9}\Big[ \frac{2\times 10^{-4} }{5 }+ \frac{1\times 10^{-4} }{5\sqrt{2} } } - \frac{2\times 10^{-4} }{5\sqrt{2} }+ \frac{1\times 10^{-4} }{5}}} \Big]$

             $= 9\times 10^{3}\Big[ \frac{(5\sqrt{2})2\times 10^{-4}+5\times 10^{-4} - 10\times 10^{-4} -5\sqrt{2}\times 10^{-4}}{5 \times 5\sqrt{2}}\Big]$  

             $= 0.9\Big[ \frac{10\sqrt{2}+5 - 10 -5\sqrt{2}}{5 \times 5\sqrt{2}}\Big]$

             $= 0.9\Big[ \frac{5\sqrt{2} - 5 }{5 \times 5\sqrt{2}}\Big]= 0.9\Big[ \frac{\sqrt{2} - 1 }{ 5\sqrt{2}}\Big]$

             $\approx 0.053J$

Work done in moving a charge from C to D is 0.053J