The magnitude of electric field at distance r from an infinitely thin rod having a linear charge density λ , is (use Gauss's law)

Question

The magnitude of electric field at distance r from an infinitely thin rod having a linear charge density λ , is (use Gauss's law) (A) E=(λ/2 π ε0 r) (B) E=(2 λ/π ε0 r) (C) E=(2 λ/4 π ε0 r) (D) E=(4 λ/π ε0 r)

Tardigrade Admin · Accepted Answer

Gauss theorem states that the total electric flux

(ϕ)

through any closed surface is equal to

\frac{1}{ε _{0}}

times the 'net' charge

q

enclosed by the surface.

ϕ_{z} = \int E \cdot d A = \frac{Q}{ε _{0}}

Taking cylindrical Gaussian surface of radius

r

, height h curved surface

= 2 π r h

.
Electric flux through it

= E \times 2 π r h

Charge enclosed

= λh

∴

From Gauss theorem,
Charge

= ε_{0} \times

(Electric flux)

\Rightarrow ε_{0} E \times 2 π r h = λh

\Rightarrow E = \frac{1}{2 π ε _{0}} \cdot \frac{λ}{r} = \frac{λ}{2 π ε _{0} r}

Q. The magnitude of electric field at distance $r$ from an infinitely thin rod having a linear charge density $λ,$ is (use Gauss's law)

$E = \frac{λ}{2 π ε _{0} r}$

$E = \frac{2 λ}{π ε _{0} r}$

$E = \frac{2 λ}{4 π ε _{0} r}$

$E = \frac{4 λ}{π ε _{0} r}$

Q. The magnitude of electric field at distance r from an infinitely thin rod having a linear charge density λ, is (use Gauss's law)

E=2πε0​rλ​

E=πε0​r2λ​

E=4πε0​r2λ​

E=πε0​r4λ​

Q. The magnitude of electric field at distance $r$ from an infinitely thin rod having a linear charge density $λ,$ is (use Gauss's law)

$E = \frac{λ}{2 π ε _{0} r}$

$E = \frac{2 λ}{π ε _{0} r}$

$E = \frac{2 λ}{4 π ε _{0} r}$

$E = \frac{4 λ}{π ε _{0} r}$