Electric Field Intensity & Calculations - NEET Physics Class 12 Free MCQ

Let net electric is zero at point P at a distance r from the 2mc charge.
Electric field at point P due to charge 2mc i sin direction of 8mc and its value is

Electric field at point P due to charge 8mc isin direction of 2mc and its value is

for net field to be zero

taking root on both sides

Side AP =  = 0.5m Electric Field due to charge at A on point P = Elecric Field due to charge at B on 

point P 

Where So force due to this field on charge P

F = qE =  4 x 10^-6 x 115.2 x 10³ F = 0.46N

Since this force in the direction of positive x axis so angle  = 0^o

By definition of Electric Field,

The plate which connected to high voltage generator induces negative charge on ball which causes attraction. When the ball strikes the positive plate, charge distribution again takes place that is the bass becomes positive and repulsion takes place. When it strikes the plate which connected to earth than its charge goes to earth and again it will be attracted towards positive plate. Hence the ball swings backward and forward hitting each plate in turn.

Electric field is the force per unit charge hence, the correct expression is: E=F/Q.

Let q charge is situated at themidposition of the lineAB.The distance between AB is x. A and B be the positions of charges Q and Q respectively.

E = F/q = KQ/r²

Properties of Electric Field Lines. At every point on an electric field line, the tangent represents the direction of electric field. If two field lines cross each other at the point, then there would be two possible directions for electric field which is physically impossible.

Electric Field Intensity E = F\q

A small negatively charged body placed at X would be pushed to the right

The negative charge feels a force opposite to the direction of field.

 Other cases are not possible from properties of electric lines of force.

E = F/q

To find the electric field at the origin, we can use the formula:

• E = F / q

Where:

• E = electric field (N/C)
• F = force experienced by the test charge (N)
• q = magnitude of the test charge (C)

Given:

• F = 9 x 10^-4 N
• q = 3 nC = 3 x 10^-9 C

Now, substituting the values into the formula:

• E = (9 x 10^-4 N) / (3 x 10^-9 C)
• E = 3 x 10⁵ N/C

Thus, the electric field at the origin is 3 x 10⁵ N/C.

Conventionally, field lines originate from a positive point charge and terminate at negative charge

When a tangent is drawn at any point on field line then that tangent gives the direction of electric field at that point

Only alpha rays are moving with small velocity and having charge so they will be affected by electric field.
X Rays and Gamma Rays are electromagnetic radiations. They do not carry electric charge while neutron is a charge less particle.

Uniform field lines imply that every point in space has same magnitude and direction of Electric Field. It is represented by parallel and equally spaced arrows in the direction of electric field

Electron is negatively charged and force is always opposite to the direction of electric field. So the when the force acts along South and opposite to the velocity which means the electron slows down.

Given:

Charge, q = 1.6 x 10^-19 C 

Mass, m = 3.2 x 10^-27 kg 

distance of null point from small charge, d=distance between two charges, q2=bigger charge, q1=small charge. +1 for like charges and -1 for unlike charges x 

Superposition of Electrostatic Force given by Coulomb’s Law for each of the charge