Two plates are at potentials -10, V and +30, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The electric field $E$ between two parallel plates is given by the formula $E = \frac{\Delta V}{d}$,where $\Delta V$ is the potential difference a

Vedclass · Accepted Answer

$2000$

Vedclass · Answer

(A) The electric field $E$ between two parallel plates is given by the formula $E = \frac{\Delta V}{d}$,where $\Delta V$ is the potential difference and $d$ is the separation distance.Given: Potential difference $\Delta V = V_2 - V_1 = 30\, V - (-10\, V) = 40\, V$.Separation $d = 2\, cm = 2 	imes 10^{-2}\, m$.Substituting these values into the formula:$E = \frac{40}{2 	imes 10^{-2}} = 20 	imes 10^2 = 2000\, V/m$.Therefore,the electric field is $2000\, V/m$.

Two plates are at potentials $-10\, V$ and $+30\, V$. If the separation between the plates is $2\, cm$,the electric field between them is.......$V/m$.

Similar Questions

Let $V$ and $E$ be the potential and electric field intensity at a point,respectively. Then:

Two plates are $20 \ cm$ apart and a potential difference of $10 \ V$ is applied between them. The electric field between the plates is . . . . . . . (in $Vm^{-1}$)

Electric potential at any point is given by $V = -5x + 3y + \sqrt{15}z$. The magnitude of the electric field is:

Obtain the relation between electric field and electric potential.

Variation in electric potential is maximum if one goes

Vedclass Test Series

Exam Paper Generator

Online Exam Module