Q. 413.5( 6 Votes )

# A 10 cm long rod carries a charge of +50 μC distributed uniformly along its length. Find the magnitude of the electric field at a point 10 cm from both the ends of the rod.

Answer :

**Given:**Length of the rod : L = 10cm = 0.1m

Charge on the rod; q = +50 μC= 50× 10

^{-6}C

Here, C is 10 cm = 0.1 m away from both ends of the rod AB.

Distance between C and centre of AB : r**Formula used:**From Pythagoras Theorem,

r

^{2}+(L/2)

^{2}=(BC)

^{2}

∴ r^{2}= [0.1]^{2}- [0.05]^{2}

∴ r^{2=} 7.5×10^{-3}

∴r=√ (7.5×10^{-3})

∴r=0.0866 m

Now we know that, Electric field at a point on the perpendicular bisector of a uniformly charged rod is:

Here, k is a constant and k= =9× 10^{9} Nm^{2}C^{-2} . q is the point charge, L is the length of the rod and Q is the magnitude of the charge.

Substituting we get,

Hence, electric field at a point 10 cm away from the ends of the rod is 5.2 × 10^{7} NC^{-1}.

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The electric force experienced by a charge of 1.0 × 10^{–6} C is 1.5 × 10^{–3} N. Find the magnitude of the electric field at the position of the charge.

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HC Verma - Concepts of Physics Part 2