Normal Intensity

Electricity and Magnetism Level 3

A system of stationary charges is symmetric relative to an axis $OO_{1}$ . At a great distance from these charges, at point $A$ on the axis, the electric field is $E_{1}=100\ \frac{V}{m}$ ; at point $B$ it is $E_{2}=99\ \frac{V}{m}$ .

The distance between $A$ and $B$ is $L=1m$ . Let's move from point $A$ to a point $r=1cm$ away from the axis, perpendicular to it. What is the perpendicular component of the electric field at this point?

The answer is 0.005.

2 solutions

Steven Chase
Nov 11, 2019

We are not told the exact charge configuration, so I will choose the simplest possible one: a single point charge at the origin. From the problem statement, we have the following two equations:

$\frac{\alpha}{D^2} = 100 \\ \frac{\alpha}{(D+1)^2} = 99$

This yields:

$D = \frac{1}{\sqrt{100/99} - 1} \approx 198.4987$

Then we can solve for $\alpha$ :

$\alpha = 100 D^2$

Then the normal field for the final configuration is:

$E_x = \frac{\alpha}{D^2 + r^2} \frac{r}{ \sqrt{D^2 + r^2}} \approx 0.0050378$

Sir, by solving this using flux considerations (imagining a cylinder with length $L$ and radius $r$ with endpoints at $A$ and $B$ ) we get the answer to be EXACTLY 0.005.

Aaghaz Mahajan - 1 year, 7 months ago

Perhaps you could post your solution

Steven Chase - 1 year, 7 months ago

I have a bug report. I don't know how exactly to post one so am commenting. I answered this problem correctly using the andriod app. After answering correctly, I viewed the solution and turned off the app. On revisiting the problem, I observed that my answer is marked as incorrect. This isn't the first time this is happening while I use the andriod app. In fact, there is a lot more that seems not okay to me, but it would take a longer comment to articulate it all.

On another note, nice problem. I solved it by considering a system of two positive point charges on either side of the axis at an arbitrary distance $d$ from the axis. By appropriately using the binomial theorem given that we are dealing with 'great distances', i approximated the electric field and finally got the answer as precisely 0.005 N/C.

Karan Chatrath - 1 year, 7 months ago

A Former Brilliant Member
Nov 11, 2019

From large distances, the charge distribution can be considered as a point charge placed at the centre of symmetry, which is on the axis $OO_1$ . From the given informations we get the distance of the point $A$ from this centre as $198 m$ . Hence the required component of electric field is $\dfrac{.01}{198}\times {100}=\boxed {.005\dfrac{V}{m}}(approx)$

The solution is wrong.....the answer comes out to be EXACTLY .005 not approx. And how did you find the distance till point A. It is already mentioned that A and B are both at a really great distance from the charge distribution.

Aaghaz Mahajan - 1 year, 7 months ago

At a really great distance from a charge distributed within a finite region, the electric field strength is $0$ . Applying Coulumb's law one can get the distance upto the first ordered approximation.

A Former Brilliant Member - 1 year, 7 months ago

Normal Intensity

The answer is 0.005.

2 solutions

0 pending reports