Question

6. There is an electric field close to the surface of Earth. This field points toward the surface and has a magnitude of about 1.5 x 102 N/C. A charge moves perpendicularly toward the surface of Earth through a distance of 439 m, the height of the Sears Tower in Chicago, Illinois. During this trip, the electric potential energy of the charge decreases by 3.7 x 10s J. a. What is the charge on the moving paride? b. What is the potential difference between the top of the Sears Tower and the ground? c. What is the electric potential of the charge at its final position?

Answer 1

(a) SInce electric field is directed towards the ground and on moving towards the ground the charge losses its potential energy i.e it gains kinetic energy.It means electrostatic force on charge is also towards the ground i.e in the same direction of field so we can say that charge is positive.

         now,

Work done by field on the charge=increase in K.E.=Decrease in P.E.

=>$F\times d=3.7\times 10^{-8}$

=>$(qE)\times d=3.7\times 10^{-8}$

$\Rightarrow q=\frac{3.7\times 10^{-8}}{Ed}$

where d= distance covered by charge=439m

$\Rightarrow q=\frac{3.7\times 10^{-8}}{1.5\times 10^{2}\times 439}$

$\mathbf{\Rightarrow q=5.62\times 10^{-13}C}$

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(b) Work done by field=change in potential energy

$q.\Delta V=3.7\times 10^{-8}J$

$\Delta V=\frac{3.7\times 10^{-8}}{q}$

$\Rightarrow \Delta V=\frac{3.7\times 10^{-8}}{5.62\times 10^{-13}}$

$\mathbf{\Rightarrow \Delta V=6.58\times 10^{4}V}$

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(C)Finally charged particle falls on earth and becomes neutralized so finally its potential becomes zero because potential of earth is also zero.