The dimensional formula of electric flux isA. `[M^(1)L^(1)T^(-2)]`B. `[M^(1)L^(3)T^(-3)A^(-1)]`C. `[M^(2)L^(2)T^(-2)A^(-2)]`D. `[M^(1)L^(-3)T^(3)A^(1)]`

The dimensional formula of electric flux isA. `[M^(1)L^(1)T^(-2)]`B. `

1.	The dimensional formula of electric flux isA. `[M^(1)L^(1)T^(-2)]`B. `[M^(1)L^(3)T^(-3)A^(-1)]`C. `[M^(2)L^(2)T^(-2)A^(-2)]`D. `[M^(1)L^(-3)T^(3)A^(1)]`
Answer» Correct Answer - B Electric flux, `phi=intvecE*vecs` The dimension of `phi` = Dimension of E `xx` dimension of s `[phi]=[M^(1)L^(1)T^(-2)][AT]^(-1)[L^(2)]=[M^(1)L^(3)T^(-3)A^(-1)]`

Discussion

No Comment Found

Related InterviewSolutions

What is the net flux of the uniform electric field of the above question through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
An electric dipole consists of charges `pm 2.0 xx 10^(8) C` separated by a distance of `2.0 xx 10^(-3)` m. It is placed near a long line charge of linear charge density `4.0 xx 10^(-4) C m^(-1)` as shown in figure (30-W4), Such that the negative charge is at a distance of `2.0 cm from the line charge. Find the force acting on the dipole. A. 7.2 N towards the line chargeB. 6.6 N away from the line chargeC. 0.6 N away from the line chargeD. 0.6 N towards the line charge.
A particle of mass `10^(-3)`kg and charge `5muC` is thrown at a speed of `20ms^(-1)` against a uniform electric field of strength `2xx10^(5)NC^(-1)`. The distance travelled by particle before coming to rest isA. 0.1 mB. 0.2 mC. 0.3 mD. 0.4 m
A uniform electric field `E=2xx10^(3)NC^(-1)` is acting along the positive x-axis. The flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane isA. `20NC^(-1)m^(2)`B. `30NC^(-1)m^(2)`C. `10NC^(-1)m^(2)`D. `40NC^(-1)m^(2)`
The electrostatic attracting froce on a small sphere of charge `0.2muC` due to another small sphere of charge `-0.4muC` in air is 0.4N. The distance between the two spheres isA. `43.2xx10^(-6)m`B. `42.4xx10^(-3)m`C. `18.1xx10^(-3)m`D. `19.2xx10^(-6)m`
A point charge `4muC` is at the centre of a cubic Gaussian surface 10 cm on edge. Net electric flux through the surfaceisA. `2.5xx10^(5)Nm^(2)C^(-1)`B. `4.5xx10^(5)Nm^(2)C^(-1)`C. `4.5xx10^(6)Nm^(2)C^(-1)`D. `2.5xx10^(6)Nm^(2)C^(-1)`
A point positive charge is brought near an isolated conducting sphere as shown in figure the electric field is best given by A. Figure (i)B. Figure (ii)C. Figure (iii)D. Figure (iv)
A coin is made up of Al and weighs 0.75 g. It has a square shape and its diagonal measures 17 mm. It is electrically neutral and contains equal amounts of positive and negative charges. The magnitude of these charges is (Atomic mass of Al = 26.98 g)A. `3.47xx10^(4)C`B. `3.47xx10^(2)C`C. `1.67xx10^(20)C`D. `1.67xx10^(22)C`
In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is , however, not constant but increases uniformly along the positive z-direction. At the rate of `10^(5)NC^(-1)m^(-1)` . What are the force and torque experienced by system having a total dipole moment equal to `10^(-7)`Cm in the negative z-direction?
A point charge causes an electric flux of `-1.0xx10^(3) N m^(2)//C` to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface ? (b) What is the is the value of the point charge ?

The dimensional formula of electric flux isA. `[M^(1)L^(1)T^(-2)]`B. `[M^(1)L^(3)T^(-3)A^(-1)]`C. `[M^(2)L^(2)T^(-2)A^(-2)]`D. `[M^(1)L^(-3)T^(3)A^(1)]`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment