(A) Given: Distance $d = 5 \, mm = 5 \times 10^{-3} \, m$,Potential difference $V = 50 \, V$,Mass $m = 10^{-15} \, kg$,Charge $q = 10^{-11} \, C$.Elec

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(A) Given: Distance $d = 5 \, mm = 5 \times 10^{-3} \, m$,Potential difference $V = 50 \, V$,Mass $m = 10^{-15} \, kg$,Charge $q = 10^{-11} \, C$.Electric field $E = \frac{V}{d} = \frac{50}{5 \times 10^{-3}} = 10^4 \, V/m$.The force on the particle is $F = qE$.Using Newton's second law,$F = ma$,so $a = \frac{qE}{m}$.Substituting the values: $a = \frac{10^{-11} \times 10^4}{10^{-15}} = \frac{10^{-7}}{10^{-15}} = 10^8 \, m/s^2$.

Class 12 Physics Electric Potential and Capacitance Relation between Electric Field and Potential and Potential Gradient

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Two parallel plates are separated by a distance of $5 \, mm$. The potential difference between them is $50 \, V$. $A$ particle of mass $10^{-15} \, kg$ and charge $10^{-11} \, C$ enters the region with a velocity of $10^7 \, m/s$. The acceleration of the particle will be:

A
$10^8 \, m/s^2$
B
$5 \times 10^5 \, m/s^2$
C
$10^5 \, m/s^2$
D
$2 \times 10^3 \, m/s^2$

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Class 12

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Electric Potential and Capacitance

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Relation between Electric Field and Potential and Potential Gradient

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.......$V/m$.

Easy

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In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (in $V$)

DifficultAP EAMCET 2022

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The electric field at a distance $x$ from the origin is given by $E = 100/x^2$. The potential difference between the points at $x = 10 \, m$ and $x = 20 \, m$ is ...... $V$.

Difficult

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The figure shows the electric potential $V$ as a function of distance through five regions on the $x$-axis. Which of the following is true for the electric field $E$ in these regions?

Easy

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If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$,the electric field (in $N/C$) at point $(1, 1, 0)$ is:

EasyAIPMT 2015

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Two parallel plates are separated by a distance of $5 \, mm$. The potential difference between them is $50 \, V$. $A$ particle of mass $10^{-15} \, kg$ and charge $10^{-11} \, C$ enters the region with a velocity of $10^7 \, m/s$. The acceleration of the particle will be:

Similar Questions

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.......$V/m$.

In a space having electric field $\vec{E}=A(x \hat{i}+y \hat{j})$,the potential at a point $(10 \ m, 20 \ m)$ is zero. Then the potential at the origin is $\left[A=10 \ Vm^{-2}\right]$. (in $V$)

The electric field at a distance $x$ from the origin is given by $E = 100/x^2$. The potential difference between the points at $x = 10 \, m$ and $x = 20 \, m$ is ...... $V$.

The figure shows the electric potential $V$ as a function of distance through five regions on the $x$-axis. Which of the following is true for the electric field $E$ in these regions?

If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$,the electric field (in $N/C$) at point $(1, 1, 0)$ is:

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