Physics Test 123

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Physics Test 123

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Weekly Quiz Competition

Question 1
4 / -1

Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that

A
E_A=E_C>E_B

B
E_A=E_B=E_C

C
E_A>E_B>E_C

D
E_A=E_C=E_B

Solution

Electric field is stronger where electric field lines are closer as depicted in the figure the field lines are almost having same distance near point A and C thus they are equal and near point B they are far thus indicating weak field at B .

At A and C, electric lines are equally spaced and dense that's why EA = EC > EB
Question 2
4 / -1

When the separation between two charges is increased,the electric potential energy of the charges

A
increases

B
decreases

C
remains the same

D
may increase or decrease
Question 3
4 / -1

An electric dipole is placed in a uniform electric field.The net electric force on the dipole

A
is always zero

B
depends on the orientation of the dipole

C
can never be zero

D
depends on the strength of the dipole
Question 4
4 / -1

The electric field in a region is directed outward and is proportional to the distance r from the origin.Taking the electric potential at the origin to be zero,

A
it is uniform in the region

B
it is proportional to r

C
it is proportional to r²

D
it increases as one goes away from the origin
Question 5
4 / -1

A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is 10 V/m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become

A
10 V/m

B
20 V/m

C
5 V/m

D
zero
Question 6
4 / -1

Electric charge are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10cm surrounding the total charge is 25V-m The flux over a concentric sphere of radius 25cm will be

A
25V–m

B
50V–m

C
100V–m

D
200V–m
Question 7
4 / -1

100V–m 4. A charge q is placed at the centre of the open end of a cylindrical vessel in figure. The flux of the electric field through the surface of the vessel is

A
zero

B
q/∈₀

C
q/2∈₀

D
2q/∈₀
Question 8
4 / -1

Mark the correct options

A
Gauss’s law is valid only for symmetrical charge distributions

B
Gauss’s law is valid only for charges placed in vaccum

C
The electric field calculated by Gauss’s law is the field due to the charges inside the Gaussian surface

D
The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface
Question 9
4 / -1

Figure shows a closed surface which intersects a conducting sphere. If a positively charged is placed at the point P, the flux of the electric field through the closed surface

A
will remain zero

B
will become positive

C
will become negative

D
will become undefined
Question 10
4 / -1

Two capacitors each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be

A
2 C and 2 V

B
C/2 and V/2

C
2 C and V/2

D
C/2 and 2 V

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Re-Attempt Weekly Quiz Competition

Physics Test 123

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Physics Test 123

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Rank

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SHARING IS CARING

Question 1

EA=EC>EB

EA=EB=EC

EA>EB>EC

EA=EC=EB

Solution

Question 2

increases

decreases

remains the same

may increase or decrease

Question 3

is always zero

depends on the orientation of the dipole

can never be zero

depends on the strength of the dipole

Question 4

it is uniform in the region

it is proportional to r

it is proportional to r2

it increases as one goes away from the origin

Question 5

10 V/m

20 V/m

5 V/m

zero

Question 6

25V–m

50V–m

100V–m

200V–m

Question 7

zero

q/∈0

q/2∈0

2q/∈0

Question 8

Gauss’s law is valid only for symmetrical charge distributions

Gauss’s law is valid only for charges placed in vaccum

The electric field calculated by Gauss’s law is the field due to the charges inside the Gaussian surface

The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface

Question 9

will remain zero

will become positive

will become negative

will become undefined

Question 10

2 C and 2 V

C/2 and V/2

2 C and V/2

C/2 and 2 V

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E_A=E_C>E_B

E_A=E_B=E_C

E_A>E_B>E_C

E_A=E_C=E_B

it is proportional to r²

q/∈₀

q/2∈₀

2q/∈₀