Test: Electrostatics - NEET MCQ

A stationary charge doesn't produce a magnetic field , it only produces an electric field.

When -q is displaced left or right of the force is increased so -q move towards that force hence will not perform SHM but when its displaced to AB it will perform SHM because it will return to its mean position.

Earlier force between two particles is 
F=kq^1​q²​​/ r²
When distance is halved force becomes 
F′= kq₁​q²/(r/2)² ​​=4kq₁​q₂/r²​​=4F

d=8.94 cm or
d=8.94x10^-2

The force between q & Q-q is given by :-
F = kq(Q-q)/r²
As it is given that force between these two is Maximum....so it will be max. when dF/dq is zero
i.e, dF/dq = {k(Q-2q)}/r² = 0
so, Q = 2q
Q/q = 2/1

There is a condition of weightlessness in a satellite.
Therefore, mg=0
Due to electrostatic force of repulsion between the balls, the string would become horizontal.
Therefore, angle between the string =180
 

x=l/[ √(q₂/q₁)-1]
=3/√[(4x10^-6/1x10^-6)-1]
= (3/2)-1
=3
So, 3m away from -1x10^-6C

Where E=0 this point is called neutral point.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi€
for 4q charge
let at ''a'' distance we get E=0 which is from q charge
so distance from 4q of 'a' point is 30-a
electric field by 4q charge on a is
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.

Electric force on a charge q placed in a region o electric field intensity is E and it is given by F = qE.
In this case, F = 100 N and q = 2 C.
So, E=F/q​=100N/2C​=50 N/C.
Hence, the value of field intensity will be 50 N/C.

As the components of electric field intensity at diagonal are equal in magnitude and opposite in direction, thus the intensity of electric field will be zero.
Hence the correct answer would be option A.

We know that, qE=mg
So,E=mg/q
E=9.1x10^-31x9.8x10¹⁹/1.6
E=9.1x9.8x10^-12/1.6
E=5.6x10^-11

Field intensity is proportional to the inverse of the square of the distance separating the point charges:

F = k*q1*q2/d^2

Since k, q1, and q2 are assumed to be constant, their product can be combined to form a single constant:

K = k*q1*q2, so

F = K/d^2 (and by algebraic manipulation we have K=F*d^2).

Therefore we know that

F(at d=4) = K/4^2, and

F(at d=2) = K/2^2.

Since K is a constant, we can equate K=F*d^2 for each case:

K=F(at d=4)*4^2 = F(at d=2)*2^2

so

F(at d=2) = (F(at d=4)*4^2)/2^2

F(at d=2) = 200N/C * (4^2)/(2^2) = 200N/C *16/4 = 800N/C.

At point P, 
E=E₁​−E₂​ rightward
⟹E=(kq/x^2​)−(kq/(l−x)²​)   where k=1/4πϵo​ 
⟹E=kq (l(l−2x)​/x²(l−x)²) 
From graph we can see that it can't be straight line and E=0 at x=l/2​
At x=0 ,,E→∞  and at x=l, E→−∞
Hence, from the shown graphs the correct answer is (D).
 

Due to verifying electric field, it experiences an verifying force :-
F=QE=QE_0​sinωt
at maximum amplitude A, it experience a maximum force of:-
F_max​=QE_0​
also, Restoring force in SHM is given by: - F=mω²x
for amplitude, x=A OR,
mω²A=QE_0​
⇒A= QE₀/mω²

Let"r" be the distance from the charge q where Q is in equilibrium .Total Force acting on Charge q and 4q:
F=kqQ/r² + k4qQ/(l-r)²
For Q to be in equilibrium , F should be equated to zero.
kqQ/r² + k4qQ/(l-r)²=0
(l-r)²=4r²
⇒l-r=2r
⇒l=3r
⇒r=l/3
Taking the third charge to be -Q (say) and then on applying the condition of equilibrium on + q charge
kQ/(L/3)² =k(4q)/L²
9kQ/L²=4kq/L²
9Q=4q
Q=4q/9
Therefore a point charge -4q/9 should be placed at a distance of L/3 rightwards from the point charge +q on the line joining the 2 charges.

The charges will be balanced by their counterparts on the opposite side. So, eventually the charges remaining will be 2q and b and 3q on D.
Since the charge distribution is asymmetrical, the net force on charge would be skewed towards D.
 

Force on particle=F=qE
Hence, acceleration of particle=a=F/m​=qE/m​
Initial speed=u=0
Hence, final velocity=v=u+at=qEt/m​
Kinetic energy=K=(1/²)​mv2=(1/2)​m(qEt/m​)2
⟹K=E²q²t²​/2m

Consider the attached free body diagram for the given scenario.
From coulomb's law,
∣F1​∣=∣F2​∣= qQ /4πε_o​r²​
In y-direction, forces cancel, F_y​=0
In x-direction, forces add up. Also, it is directed in +ve x direction if the position of −q is along −ve x axis and in −ve x direction if the position of −q is along +ve x axis.
Hence, 
F_x​=2qQcosθ​ i /4πε_o​r²    ∀ x<0
F_x​=2qQcosθ​ i /4πε_o​r²    ∀ x≥0
But cosθ =x/r
       r= √x²+d^2​
Substituting in force equation, we get
F_x​=2qQx​ i /4πε_o​( x²+d²​)^3/2  ∀ x<0
F_x​=2qQx i /4πε_o​( x²+d^2​)^3/2  ∀ x≥0
 
Now, a=F/m​ by Newton's second law. It is positive when x is negative and negative when x is positive. This is satisfied only in option (B).
 

Electric field due to a charged ring in given by: -
at point p
∣​E∣​= KQx​ /(R²+x²)^3/2        Q =λ(2πr)
at a large distance, x≫R, so :- R2+x2≃x2
⇒∣​E∣​= K&x​/(x²)^3/2=KQ/x²​=Eαx⁻²
so at a large distance, the ring behaves as a point particle.

If we fill the gap then there will be no electric field at the center of ring
This signifies that electric field due to rest ring is equal to electric field due to gap and opposite in direction
so electric field due to gap= Kq/R²​
q=Qd/2πR​
thus electric field E=KQd​/ R³
Hence E will be inversely proportional to R³  and direction will be toward the gap.
 

The angle suspended by the finite line charge at P=60°
So, the resultant electric field due to line charge will be at 60/2⇒30° since we can assume the charge concentrated at the centre of finite line charge.

Electric force F=qE
given, F=3000N,q=3C
∴E=F/q​=3000/3​=1000 V/m
Potential difference, V=Ed where d=0.01m
∴V=1000×0.01=10V

Electric potential is a pure Scalar quantity, The reason is as follows.
The Electric Potential is defined as the amount of work-done per unit positive charge to bring from infinity to that point under the influence of the primary charge only.
U=W/q
And work done is defined as the dot product of force and displacement which is a scalar quantity.
W=F.S
Thus Electric potential is a scalar quantity.