Question

a wire is bent into a semicircular arc of radius R the rod has a uniform linear charge distribution Lambda. the potential at the centre is

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Answer 1

Given that,

Radius = R

The rod has a uniform linear charge distribution Lambda.

The charge is

$Q=\lambda \pi R$

Where, Q = charge

R = radius

We need to calculate the potential at the center

Using formula of potential

$V=\dfrac{kQ}{R}$

Put the value of Q in to the formula

$V=\dfrac{\lambda\pi R}{4\pi\epsilon_{0}R}$

$V=\dfrac{\lambda}{4\epsilon_{0}}$

Hence, The potential at the center is $\dfrac{\lambda}{4\epsilon_{0}}$

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Topic : potential

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Answer 2

Answer:

The potential at the center is V=λ/4ε0

Explanation:

When we get the distance of the centre from each vertex of the triangle then we can find the electric potential due to each charge placed at the vertices of the triangle. The equivalent electric potential at the centre will be the sum of electric potential due to each charge placed at vertices of the triangle. total potential at the center V=4V1=ks4 q.

Radius = R

The rod has a uniform linear charge distribution Lambda.

The charge is

Q= λπR

Where, Q = charge

R = radius

We need to calculate the potential at the center

Using formula of potential

V=KQ/R

Put the value of Q in to the formula

V=λπR/4πε0R

V=λ/4ε0

The potential at the center is V=λ/4ε0

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