Question

A thin plastic rod of length 2.6 m is rubbed all over with wool, and acquires a charge of 98 nC, distributed uniformly over its surface. Calculate the magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod. Do the calculation two ways, first using the exact formula for a rod of any length, and second using the approximate formula for a long rod.

Answer 1

By exact formula

5076.59N/C

And by approximation formula

5218.93N/C

Step-by-step explanation:

We are given that

Length of rod,L=2.6 m

Charge,q=98nC=
`98* 10^(-9) C`

`1nC=10^(-9) C`

a=13 cm=0.13 m

1 m=100 cm

By exact formula

The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=
`(kq)/(a)* \frac{1}{\sqrt{a^2+(L^2)/(4)}}`

Where k=
`9* 10^9`

Using the formula

The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=
`(9* 10^9* 98* 10^(-9))/(0.13)* \frac{1}{\sqrt{(0.13)^2+((2.6)^2)/(4)}}=5076.59N/C`

In approximation formula

a<<L

`a^2+((L)/(2))^2=(L^2)/(4)`

Therefore,the magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=
`(kq)/(a)* \frac{1}{\sqrt{(L^2)/(4)}}`

The magnitude of the electric field due to the rod at a location 13 cm from the midpoint of the rod=
`(9* 10^9* 98* 10^(-9))/(0.13)* \frac{1}{\sqrt{((2.6)^2)/(4)}}=5218.93N/C`