An infinite thin non-conducting sheet has uniform surface charge density σ . Find the potential difference VA-VB between points A and B as shown in the figure. The line AB makes an angle of 37° with the normal to the sheet. (sin 3 7 ° = (3/5))

Question

An infinite thin non-conducting sheet has uniform surface charge density σ . Find the potential difference VA-VB between points A and B as shown in the figure. The line AB makes an angle of 37° with the normal to the sheet. (sin 3 7 ° = (3/5)) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-llcppze90c89.png" /> (A) (2 σ d/5 ϵ 0) (B) (- 2 σ d/5 ϵ 0) (C) (3/10)(σ d/ϵ 0) (D) (- 3 σ d/10 ϵ 0)

Tardigrade Admin · Accepted Answer

Vf-Vi=- overset arrow E. overset arrow S (for uniform field) VB-VA=-(σ /2 ϵ 0)dcos 37° =(- 2 σ d/5 ϵ 0) ⇒ VA-VB=(2 σ d/5 ϵ 0)

Q. An infinite thin non-conducting sheet has uniform surface charge density $σ$ . Find the potential difference $V_{A} - V_{B}$ between points $A$ and $B$ as shown in the figure. The line $A B$ makes an angle of $3 7^{\circ}$ with the normal to the sheet. $(s in 3 7^{\circ} = \frac{3}{5})$

$\frac{2 σ d}{5 ϵ _{0}}$

$\frac{- 2 σ d}{5 ϵ _{0}}$

$\frac{3}{10} \frac{σ d}{ϵ _{0}}$

$\frac{- 3 σ d}{10 ϵ _{0}}$

$V_{f} - V_{i} = - E \to . S \to$ (for uniform field)
$V_{B} - V_{A} = - \frac{σ}{2 ϵ _{0}} d cos 3 7^{\circ} = \frac{- 2 σ d}{5 ϵ _{0}}$
$\Rightarrow V_{A} - V_{B} = \frac{2 σ d}{5 ϵ _{0}}$

Q. An infinite thin non-conducting sheet has uniform surface charge density σ . Find the potential difference VA​−VB​ between points A and B as shown in the figure. The line AB makes an angle of 37∘ with the normal to the sheet. (sin37∘=53​)

5ϵ0​2σd​

5ϵ0​−2σd​

103​ϵ0​σd​

10ϵ0​−3σd​