IIT JEE , AIEEE Forum - Physics , Electricity

iitaspirant1990

1 .A ROD OF LENGTH L LIES ALONG THE X-AXIS WITH ITS LEFT END AT THE ORIGIN .IT HAS A NONUNIFORM CHARGE DENSITY LAMDA = ALFA * x , WHERE ALFA IS A POSITIVE CONSTANT.CALCULATE THE ELECTRIC POTENTIAL AT POINT A WHERE x = -d.

2.A CONE MADE OF INSULATING MATERIAL HAS A TOTAL CHARGE Q SPREAD UNIFORMLY OVER IS SLOPING SURFACE.CALCULATE THE WORKDONE IN BRINGING A SMAL TEST CHARGE q FROM INFINITY TO THE APEX OF THE CONE. THE CONE HAS A SLOPE LENGTH L.

kishore.subramanian.b

for $\alpha$ x distance Q charge is distributed

for dx distance (Q/ $\alpha$ x)dx is distributed

Potential at a point x on negative x-axis due to dx is

dV = 1/4 $\pi\epsilon$ (Q/ $\alpha$ x)dx /x

= 1/4 $\pi\epsilon$ (Q/ $\alpha$ ) dx/ ${x}^{2}$

Total potential is

V = $\int_{d}^{d+L}$ 1/4 $\pi\epsilon$ (Q/ $\alpha$ ) dx/ ${x}^{2}$

=1/4 $\pi\epsilon$ QL/ $\alpha$ d(L+d)

krishna.gopal

Sorry because of some problem kishore's answer is not clear to me (i am unable to see all charechters) Here is my solution.

Consider a small element between x and x=dx from left end

So its potential

dV= k*alpha*x*dx/(x+d)

Integrate this from 0 to L and you will get the answer

So V = K*alpha*L -K*alpha*d*ln(1+L/d))

kishore.subramanian.b

sir plwase telll me which step i have gone wrong.....

Froggy

Here's how we go about it:

Considering an element of thickness dx, the charge on it = $\alpha xdx$ . Hence, the Potential due to this charge at x = -d is nothing but $V = \frac{K dq}{x+d}$ .

The potential due to the whole charge will simply be $\int^{L+d}_{d} \frac{K \alpha x}{x+d} \, dx$ where K = $\frac{1}{4 \pi \epsilon_o}$ .

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2.A CONE MADE OF INSULATING MATERIAL HAS A TOTAL CHARGE Q SPREAD UNIFORMLY OVER IS SLOPING SURFACE.CALCULATE THE WORKDONE IN BRINGING A SMAL TEST CHARGE q FROM INFINITY TO THE APEX OF THE CONE. THE CONE HAS A SLOPE LENGTH L.