Electric Charge, Coulomb's law and Electric Field Test

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Electric Charge, Coulomb's law and Electric Field Test - 12

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Weekly Quiz Competition

Question 1
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If Q =2 coloumb and force on it is F=100 newtons , Then the value of field intensity will be

A
100 N/C

B
50 N/C

C
200 N/C

D
10 N/C

Solution

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.
Question 2
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Four equal but like charge are placed at four corners of a square. The electric field intensity at the center of the square due to any one charge is E, then the resultant electric field intensity at centre of square will be

A
Zero

B
4E

C
E

D
1/2E

Solution

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.
Question 3
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If mass of the electron = 9.1 ×10^-31 Kg. Charge on the electron = 1.6 ×10^-19 coulomb and g = 9.8 m/s² . Then the intensity of the electric field required to balance the weight of an electron is

A
5.6 ×10^-9 N/C

B
5.6 ×10^-¹¹ N/C

C
5.6 ×10^-⁸ N/C

D
5.6 ×10^-7 N/C

Solution

We know that, qE=mg
So,E=mg/q
E=9.1x10^-31 x9.8x10¹⁹ /1.6
E=9.1x9.8x10^-12 /1.6
E=5.6x10^-11
Question 4
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The electric field intensity at a point situated 4 meters from a point charge is 200 N/C. If the distance is reduced to 2 meters, the field intensity will be

A
400 N/C

B
600 N/C

C
800 N/C

D
1200 N/C

Solution

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.
Question 5
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At position of (4i + 3j) by a point charge 5 ×10^-6 C placed at origin will produce electric field of

A
1240i + 1180j

B
1442i + 2080j

C
1440i + 1080j

D
2440i + 3080j

Solution
Question 6
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Two identical point charges are placed at a separation of l .P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to zero to slightly less than l . Which of the following best represents the resulting curve ?

A

B

C

D

Solution

At point P,
E=E₁ −E₂ rightward
⟹E=(kq/x² )−(kq/(l −x)² ) where k=1/4 πϵo
⟹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).
Question 7
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A particle of mass m and charge Q is placed in an electric field E which varies with time t ass E = E₀ sinwt. It will undergo simple harmonic motion of amplitude

A

B

C

D

Solution

Due to verifying electric field, it experiences an verifying force :-
F=QE=QE₀ sin ωt
at maximum amplitude A, it experience a maximum force of:-
F_max =QE₀
also, Restoring force in SHM is given by: - F=m ω² x
for amplitude, x=A OR,
m ω² A=QE₀
⇒A= QE₀ /m ω²
Question 8
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Four charges are arranged at the corners of a square ABCD, as shown. The force on +ve charge kept at the centre of the square is

A
Zero

B
Along diagonal AC

C
Along diagonal BD

D
Perpendicular to the side AB
Question 9
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Two free positive charges 4q and q are a distance l apart. What charge Q is needed to achieve equilibrium for the entire system and where should it be placed form charge q ?

A

B

C
Q = q (positive) at

D
Q = q (negative) at

Solution

Let "r "be the distance from the charge q where Q is in equilibrium .Total Force acting on Charge q and 4q:
F=kqQ/r ²+ k4qQ/(l-r)²
For Q to be in equilibrium , F should be equated to zero.
kqQ/r ²+ k4qQ/(l-r)²=0
(l-r)²=4r ²
⇒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)/L ²
9kQ/L ²=4kq/L ²
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.
Question 10
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Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be

A
Zero

B
Along OF

C
Along OC

D
None of these

Solution

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.