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Question: A rod of length \( 2.4m \) and radius \( 4.6mm \) carries a negative charge of \( 4.2 \times {10^{ -...

A rod of length 2.4m2.4m and radius 4.6mm4.6mm carries a negative charge of 4.2×10−7C4.2 \times {10^{ - 7}}C spread uniformly over its surface. The electric field near the midpoint of the rod, at a point on its surface is:
A, −8.6×105NC−1- 8.6 \times {10^5}N{C^{ - 1}}
B. 8.6×104NC−18.6 \times {10^4}N{C^{ - 1}}
C. −6.7×105NC−1- 6.7 \times {10^5}N{C^{ - 1}}
D. 6.7×10NC−16.7 \times 10N{C^{ - 1}}

Explanation

Solution

Hint: You can start the solution by describing what electric field is and how it is formed. Then move to use the equation λ=ql\lambda = \dfrac{q}{l} to find the linear charge density of the rod. Then use the equation E=λ2πε0rE = \dfrac{\lambda }{{2\pi {\varepsilon _0}r}} to reach the solution.

Complete step-by-step answer:
Before moving on to the mathematical calculations, let’s first discuss why an electric charge forms an electric field around it.
Any charge that comes in the electric field of another charge experiences an electrostatic force of attraction or repulsion.
In most theoretical questions we are given a point charge, but here a negative charge of 4.2×10−7C4.2 \times {10^{ - 7}}C is distributed on a rod of length 2.4m2.4m and a radius of 4.6mm4.6mm .
Now we know
λ=ql\lambda = \dfrac{q}{l}
Here
λ=\lambda = Linear charge density
q=q = Total charge
l=l = Length of rod
So for this problem
q=q = −4.2×10−7C- 4.2 \times {10^{ - 7}}C
l=l = 2.4m2.4m
So, λ=4.2×10−7C2.4m\lambda = \dfrac{{4.2 \times {{10}^{ - 7}}C}}{{2.4m}}
⇒λ=−1.75×10−7C/m\Rightarrow \lambda = - 1.75 \times {10^{ - 7}}C/m
We do not consider the radius or area of cross-section of the metal rod, while calculating the linear charge density.
Now for finding out the electric field on the middle point we use the equation
E=λ2πε0rE = \dfrac{\lambda }{{2\pi {\varepsilon _0}r}}
Here
λ=\lambda = Linear charge density
π=\pi = 3.14
{\varepsilon _0} = $$$${\varepsilon _0} = 8.85 \times {10^{ - 12}}F{m^{ - 1}}
r=r = Radius of the wire
E=−1.75×10−72×3.14×10−12×4.6×10−3E = \dfrac{{ - 1.75 \times {{10}^{ - 7}}}}{{2 \times 3.14 \times {{10}^{ - 12}} \times 4.6 \times {{10}^{ - 3}}}}
E=−6.67×105NC−1E = - 6.67 \times {10^5}N{C^{ - 1}}
Hence, option C is the correct choice.

Note – We discussed earlier that a charge that enters the electric field of another charge experiences a force. It is essential to clarify that an electric field is not a physical entity, it’s an imaginary space around the charge where another charge experiences a force due to it. The shape of the electric field formed due to the presence of two charges is influenced by both the charges.