Consider the equipotential diagram at the top of the next page showing two missh
ID: 1527743 • Letter: C
Question
Consider the equipotential diagram at the top of the next page showing two misshapen conductors (the shaded regions labeled I and I). One of the conductors has a net charge of +10 mu C and the other conductor has an equal and opposite charge of -10 mu C. Consider a positive charge placed at point P. Based on the equipotential diagram, will this charge naturally want to move to the left or to the right? Does the electric field point to the left or to the right? Based on your answer, which conductor, I or II, has the +10 mu C and which has the -10 mu C charge? Explain your answers. Sketch some electric field lines for this equipotential diagram Estimate the strength of the electric field at point P. Assume that one "box' of the graphing-paper grid corresponds to a length of d = 1 cm. This pair of conductors acts as a capacitor. What is the capacitance of this capacitor How much work should it take to move a single electron (charge q =- e) from region I to region II? How much total energy is stored in this capacitor? Let region I be maintained at a voltage of -3 V. If the charge on each conductor were doubled, what would the new voltage of region II be?Explanation / Answer
Given
point p is at a distance of 3 cm from conductor 1 and 4 cm from second conductor
and at point p a positive charge is placed so it would move towards second conductor
so the first one is positive charge and second one is negative charge
1 --- +10 *10^-6 C,
2---- -10 *10^-6 C
electric field points to right side
for a positive charge the field lines are radially outward and for -ve charge radially inward
and two opposite charges attract each other .
as we know that the field is like mediator for force between two charges.
g)strength of electric field at point P is
from relation V = E*d
E = DV/d
E1 = 2/3 , E2 = 2/4
E = 0.666+0.5 = 1.166 V/m
h)
capacitance C, Q = C*V ==> C = Q/V = (10*10^-6 )/(5) F = 2*10^-6 F
i ) work done W = DV*q = 5*1.6*10^-19 J = 8*10^-19 J required
j) energy stored is u = 0.5*c*v^2
= 0.5*2*10^-6*5^2 = 25*10^-6 J