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The vertical deflection of the particle at the far edge of the plate is
given s= 0.5cm=0.005cm
calculate for L from the above equation
L=1.6 cm

Let s be the vertical deflection, t be the time taken by the particle to travel between the plates
Here , u =0 , since initially there was no vertical component of velocity.
The particle in the elecric field will experience a constant force (Since, Electric field is constant.)
F = ma = -qE (Using Newton's Second Law, F = ma)
a = -qE/m (-ve sign implies here in downward direction)
Again, t =...

Two charges of same magnitude and sign are placed at a certain distance apart. The mid-point of the line joining these charges will have E =0.
When a test charged is displaced along the line towards the charges, it experiences a restoring force(which is the condition for stable equilibrium). But if the test charge is displaced along the normal of the line, the net electrostatic force pushes it...

For equilibrium to be stable, there must be a restoring force and hence all the field lines should be directed inwards towards the null point. This implies that there is a net inward flux of the electric field. But this violates Gauss's law, which states that the flux of electric field through a surface not enclosing any charge must be zero. Hence, the equilibrium of the test charge cannot be...

Given, a proton and a neutron consist of three quarks each.
And, ‘up’ quark is of charge + e, and the ‘down’ quark of charge e
Let the number of 'up' quarks be n. Therefore, the number of 'down' quarks is (3-n).
The net charge =
Now, A proton has a charge +1e
Proton will have 2 u and 1 d, i.e, uud
Similarly, neutron has a charge 0
Neutron will have 1 u and 2 d, i.e, udd

Let AB be a long thin wire of uniform linear charge density λ.
Let us consider the electric field intensity due to AB at point P at a distance h from it as shown in the figure.
The charge on a small length dx on the line AB is q which is given as q = λdx.
So, according to Coulomb’s law, the electric field at P due to this length dx is
But
⇒
This electric field at P can be resolved into...

Let the surface area of the sphere be S.
And assume that the hole is filled. For a point B, just above the hole, considering a gaussian surface passing through B, we have
Now, Since the electric field is always perpendicular to the surface of the conductor.
Using Superposition principle, ,
where is due to the hole and is due to the rest of the conductor. (Both pointing outwards, i.e...

We know that electric field inside a conductor is zero.

Therefore, a possible way to shield from the strong electrostatic fields in its environment is to enclose the instrument fully by a metallic surface.

We know, Electric field inside a conductor is zero.
Now, imagine a Gaussian surface just outside the cavity inside the conductor. Since, E=0 (Using Gauss' law), hence net charge must be zero inside the surface. Therefore, -q charge is induced on the inner side of the cavity(facing conductor B).
Now consider a Gaussian surface just outside the conductor A. The net electric field must be due to...

We know that electric field inside a conductor is zero.
Using Gauss' law, if we draw any imaginary closed surface inside the solid, net charge must be zero.(Since E= 0 inside)
Hence there cannot be any charge inside the conductor and therefore, all charge must appear on the outer surface of the conductor.

Force on a charge F=qE
but here E is varying along the Z direction.
Force can be written as,
Torque experienced =0 since both dipole and electric field are in the Z direction. The angle between dipole and the electric field is 180 degrees

(a) Wrong, because field lines must be normal to a conductor.
(b) Wrong, because field lines can only start from a positive charge. It cannot start from a negative charge,
(c) Right;
(d) Wrong, because field lines cannot intersect each other,
(e) Wrong, because electrostatic field lines cannot form closed loops.

The force due to the electric field is balancing the weight of the oil droplet.
Weight of the oil drop = Densityx Volume of the droplet x g =
Force due to the electric field = E x q
Charge on the droplet, q = No. of excess electrons x Charge of an electron =
Balancing forces:
Putting known and calculated values:

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density = .
(To note: It's independent of distance from the plate!)
Let A and B be the two plates such that:
=
= -
Therefore,
The electric field between the plates, E = =

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density = .
(To note: It's independent of distance from the plate and same everywhere!)
In the region outside the second plate,
Since both plates have the same surface charge density(in magnitude only), their electric fields are same in magnitude in this region but opposite in direction.
(E due...

We know, electric field, E, due to an infinite plate (length>>thickness) having surface charge density = .
(To note: It's independent of distance from the plate!)
In region outside first plate,
Since, both plates have same surface charge density(in magnitude only), their electric fields are same in magnitude in this region but opposite in direction.
(E due to positive plate away from it...

Given,
d = 2 cm = 0.02 m
We know, For an infinite line charge having linear charge density , the electric field at a distance d is:
The linear charge density is .

Given,
Surface charge density =
Diameter of sphere = 2.4 m Radius of sphere, r = 1.2 m
The charge on the sphere, Q= Surface charge density x Surface area of the sphere

We know, for determining the electric field at r>R for a conducting sphere, the sphere can be considered as a point charge located at its centre.
Also, electric field intensity at a point P, located at a distance r, due to net charge q is given by,
Given, r = 20 cm = 0.2 m (From the centre, not from the surface!)
Therefore, Charge on the conducting sphere is (Since flux is inwards)

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