An infinite plane of charge with σ=2 ϵ0 (C/m2) is tilted at a 37° angle to the vertical direction as shown below. Find the potential difference, VA-VB in volts, between points A and B at 5 m distance apart (where B is vertically above A ).

Question

An infinite plane of charge with σ=2 ϵ0 (C/m2) is tilted at a 37° angle to the vertical direction as shown below. Find the potential difference, VA-VB in volts, between points A and B at 5 m distance apart (where B is vertically above A ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8c61b178ceb6b3a0dcbf33ad2102b0c9-.png" /> E =(σ/2 ε0)=1 N / C ∴ VB-VA =-∫ vecE ⋅ d vecl=((σ/2 ϵ0))(5 cos 53°)=3 V

Q. An infinite plane of charge with $σ = 2 ϵ_{0} \frac{C}{m ^{2}}$ is tilted at a $3 7^{\circ}$ angle to the vertical direction as shown below. Find the potential difference, $V_{A} - V_{B}$ in volts, between points $A$ and $B$ at $5 m$ distance apart (where $B$ is vertically above $A$ ).

$E = \frac{σ}{2 ε _{0}} = 1 N / C$
$∴ V_{B} - V_{A}$
$= - \int E \cdot d l = (\frac{σ}{2 ϵ _{0}}) (5 cos 5 3^{\circ}) = 3 V$

Q. An infinite plane of charge with σ=2ϵ0​m2C​ is tilted at a 37∘ angle to the vertical direction as shown below. Find the potential difference, VA​−VB​ in volts, between points A and B at 5m distance apart (where B is vertically above A ).

E=2ε0​σ​=1N/C ∴VB​−VA​ =−∫E⋅dl=(2ϵ0​σ​)(5cos53∘)=3V

Q. An infinite plane of charge with $σ = 2 ϵ_{0} \frac{C}{m ^{2}}$ is tilted at a $3 7^{\circ}$ angle to the vertical direction as shown below. Find the potential difference, $V_{A} - V_{B}$ in volts, between points $A$ and $B$ at $5 m$ distance apart (where $B$ is vertically above $A$ ).

$E = \frac{σ}{2 ε _{0}} = 1 N / C$
$∴ V_{B} - V_{A}$
$= - \int E \cdot d l = (\frac{σ}{2 ϵ _{0}}) (5 cos 5 3^{\circ}) = 3 V$