The dimensional formula of electric field is . . . . . . .

  • A
    $M^1 L^1 T^{-3} A^{-1}$
  • B
    $M^1 L^1 T^{-2} A^{-1}$
  • C
    $M^1 L^2 T^{-3} A^{-1}$
  • D
    $M^0 L^1 T^{-3} A^{-1}$

Explore More

Similar Questions

The variation of electric field along the $Z$-axis due to a uniformly charged circular ring of radius '$a$' in the $XY$ plane is shown in the figure. The value of coordinate $M$ will be

An infinite number of electric charges, each equal to $5 \text{ nC}$ (magnitude), are placed along the $X$-axis at $x = 1 \text{ cm}, x = 2 \text{ cm}, x = 4 \text{ cm}, x = 8 \text{ cm}, \dots$ and so on. In this setup, if the consecutive charges have opposite signs, then the electric field in $\text{N/C}$ at $x = 0$ is: $\left(\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2\right)$

$A$ circular wire loop of radius $10 \ cm$ carries a total charge of $10^{-5} \ C$ distributed uniformly over its length. $A$ small length of $3.14 \times 10^{-6} \ m$ of wire is cut off. The magnitude of the electric field at the centre due to the remaining wire is:
(Assume $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \ SI$ units) (in $N \ C^{-1}$)

Find out the electric field intensity at point $A(1, 0, 2)$ due to a point charge $-20\,\mu C$ situated at point $B(0, 2, 1)$.

Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo