Answer to Question #100968 in Electricity and Magnetism for Cielo vivar

Question

What is the electric field at a distance of 3cm from 4mC of negative charge?
The electric potential energy of an object at point A is known to be 50 J. If it is released from rest at A, it gains 30 J of kinetic energy as it moves to point B. What is its potential energy at B?
How much work does a 12-V battery do in pushing 2mC of charge through a circuit containing one light bulb?
What was your electric potential relative to a metal pipe if a spark jumped 1.3 cm (0.5 in.) through dry air from your finger to the pipe?

Answer 1

Electric field is given by $Q/4\pi\epsilon r^2$

here Q=4*10^-3 C and r=3*10^-2 m

Electric field= 9*10⁹ * 4*10^-3/9*10^-4 = 4*10¹⁰ V/m

$\Delta K.E.=-\Delta U$

K.E._A - K.E._B = U_B - U_A

0-30 = U_B - 50

U_B = 20 J

Work done = charge * potential difference

Work done = 12*0.002 =0.024 J

The electric field is given by volts/distance: E= V/d . The breakdown voltage of dry air is about 3x10⁶ V/m. So solving for V we get

V=Ed

or V=(3*10⁶ V/m)*(0.013m)=39000 V