A mercury drop of water has potential V on its surface. 1000 such drops combine to form a new drop. Find the potential (in Volts) on the surface of the new drop.
Let the radius of each mercury drop be r.
If q is charge on each drop
The potential of drop
Let R be the radius of the new drop formed by combination of 1000 drops of radius r.
∴ Potential of new drop
The correct answer is: 100
If the charge in a conductor is 16C and the area of cross section is 4m2. Calculate the electric flux density.
Electric Flux density is the charge per unit area.
The formula is: D = Q / A = 16 / 4 = 4C/m2.
The number of electrons to be put on spherical conductor of radius 0.1m, to produce an electric field of 0.036N/C just above the surface is Find the value of α
The electric field at the surface of conductor
The correct answer is: 2.5
At distance of 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively. Then charge on the sphere is Find the Value of α.
The correct answer is: 1.667
Two positive point charges each of magnitude 10C are fixed at positions A and B at a separation 2d = 6m. A negatively charged particle of mass m = 90gm and charge of magnitude 10 × 10–5C is revolving in a circular path of radius 4m in the plane perpendicular to the line AB and bisecting the line AB. Neglect the effect of gravity. If the angular velocity of the particle is Then the value of α (in rad/sec) is :
Net force on –q towards the centre,
For particle to move in circle
= 400 rad/s
The correct answer is: 400
Eight point charges (can be assumed as small sphere uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube. The electric flux through square surface ABCD of the cube is Find the value of α
The electric flux through ABCD due to charges at A, B, C and D is zero.
The flux through ABCD due to any of the charges at E, F, G and H are each.
Hence the net electric flux through ABCD is
The correct answer is: 6
Figure given, shows a closed Gaussian surface in the shape of a cube of edge length 3.0m. There exists an electric field given by where x is in metres, in the region in which it lies. The net charge in coulombs enclosed by the cube is equal to α∈0. Find the value of α.
Direction of field at x = –3m is along negative x-axis. Area vector is also along same direction
Components of electric field which are constant, do not contribute in net flux in or out.
The correct answer is: 54
A square loop of side ℓ having uniform linear charge density λ is placed in xy plane as shown in the figure. There is a non-uniform electric field where a is a constant. Then resultant electric force on the loop if ℓ = 10 cm, a = 2N/C and charge density λ = 2μC/m is Find the value of α.
on BC and AD electric field is no uniform
∴ x is not constant. But on BC and AD electric field will have the same type of variation.
The correct answer is: 4
On moving a charge of 20coulombs by 2cm, 2J of work is done, then the potential difference (in Volts) between the points is :
By definition of electrostatic potential energy,
The correct answer is: 0.1
The magnitude of electric field intensity at point (2, 0, 0) due to a dipole moment, kept at origin is αk [where k = 1/π ∈0]. Find the value of α.
The dipole moment makes an angle 60° with x-axis and lies in x-y plane as shown in figure.
The electric field a point A due to dipole is
The correct answer is: 0.331