Answered: A rod with a length of l lies on the…

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Question

A rod with a length of l lies on the x-axis as shown in the picture. It carries a total charge of Q.

A) Write the integral (you do not need to solve it!) to find the electric field at the point P which lies above the line of charge on the same axis.
B) If you compute the integral, you would find that e=kq/(x(x+l))

. Say the rod carries a total charge +8nc
of and has a length of 8um. Find the force (as a vector) on a particle carrying a charge of -3uc
which is located 1.5um from the end of the rod.

Q.
0+
neod to solve it) to find the electric field at the polnt P which

Expert Solution

Step 1 Concept

Given;

A rod with length l lies on the x-axis and it carries a total charge Q.

(A) Write the integral to find the electric field at the point P which lies above the line of charge on the same axis.

Solution:

As given in the problem, the charge Q is distributed over the rod of length l. This linear charge distribution is called as a linear or line charge distribution.

$\begin{array}{rcl} I t i s d e n o t e d b y λ; \\ T h e l i n e c h a r g e d i s t r i b u t i o n i s d e f i n e d a s, \\ T h e t o t a l c h a r g e d i s t r i b u t e d p e r u n i t l e n g t h . \\ ∴ λ & = & \frac{dq}{dl} \\ Now the electric field due to liner charge distribution is given by; \\ d E & = & \frac{1}{4 {πε}_{0}} \int_{l} \frac{dq}{r^{2}} \\ d E & = & \frac{1}{4 {πε}_{0}} \int_{l} \frac{λ}{r^{2}} dl \\ Let, \frac{1}{4 {πε}_{0}} & = & k = constant \\ λ & = & \frac{dq}{dl} \\ As it is aske dto find out electric field at point P lies on x axis (same axis), \\ let us say point P at distance x from rod of length l . \\ But in the above formula r is the distance at which electric field to be measured . \\ Therefore; \\ r & = & (x + l) \\ ∴ d E & = & \frac{1}{4 {πε}_{0}} \int_{l} \frac{dq}{r^{2}} \\ b e c o m e s; \\ E & = & k \int_{0}^{l} \frac{λ}{{(x + l)}^{2}} dl \\ A n s w e r & = & T h e i n t e g r a l p a r t o f t h e e l e c t r i c f i e l d i s; E = k \int_{0}^{l} \frac{λ}{{(x + l)}^{2}} dl \end{array}$

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