Physics Test 4 - Electrostatics, Capacitance, Current Electricity
25 Questions MCQ Test JEE Main & Advanced Mock Test Series | Physics Test 4 - Electrostatics, Capacitance, Current Electricity
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Figure shows some of the electric field lines corresponding to an electric field. The figure suggests that
At point A and C, electric field lines are dense and equally spaced, so EA = EC
While at B they are far apart.
∴ EA = EC > EB
When the separation between two charges is increased, the electric potential energy of the charges
Potential energy between two charges 
Now if r increases and q1 and q2 are of same sign then
decrases. if q1 and q2 are of opposite 
increases.
Therefore depending on signs of q1 and q2 
may increase or decrease.
An electric dipole is placed in a uniform electric field. The net electric force on the dipole
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
when a charge Q is placed at the centre of the open end of a cylindrical vessel then only half of the charge will contribute to the flux,because half will lie inside the surface and half will outside the surface so flux=Q/2ε0
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 dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will
Two metal spheres of capacitances C1 & C2 carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy
Two resistances R and 2R are connected in parallel in an electric circuit. The thermal energy developed in R and 2 R are in the ratio
Thermal energy = V2t/R
In parellel arrangement V is constant
So, for a time t
Thermal energy ∝ 1/R
Ratio of their thermal energy = R₂ : R₁ = 2R : R = 2 : 1
is cut into 5 equal parts. These parts are now connected in parallel. The equivalent resistance of the combination is
The equivalent capacitance between the points A and B of five identical capacitors each of capacity C, will be
adding these in parallel will give you C/2 +C/2+C = 2C.
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
If an electron is brought towards an another electron, the electric potential energy of the system
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
E=KQ/4-------------equ-2
on dividing1by 2
E= 800
An electric dipole is placed at an angle of 30° with an electric field intensity 2 × 105 N/C. It experiences a torque equal to 4 N m. The charge on the dipole, if the dipole length is 2 cm, is
Shown in the figure is a distribution of charges. The flux of electric field due to these charges through the surface S is
Flux flowing through a body depends on the net charge enclosed inside the surface, here net charge inside is zero so flux is also zero. And the third charge is outside so it will not effect the flux.
Ten identical cells each of emf E and internal resistance r are connected in series to form a closed circuit. An ideal voltmeter connected across the three cells will read
No. of cells = 10
Emf = E
Internal resistance = r
Let I = current through the ckt
Now potential (V) of across each cell will be
V = E -Ir
We see there being no load resistance internal resistance play no effect on the observed emf
Thus the pd across 3 cells will be
E + E + E = 3Ir = 3E
A dipole consists of two particles one with charge +1µC and mass 1kg and the other with charge –1µC and mass 2kg separated by a distance of 3m. for small oscillations about its equilibrium position, the angular frequency, when placed in a uniform electric field of 20kV/m is :-
Restoring torque
τ = – PE sin θ
τ = – PE θ, for small θ
A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge –3Q, the new potential difference between the same two surface is xV, then x is :-
A potentiometer is used for the comparison of e.m.f. of two cells E1 and E2. For cell E1 the no deflection point is obtained at 20 cm and for E2 the no deflection point is obtained at 30 cm. The ratio of their e.m.f.’s is N/3 then N is:
Ratio will be equal to the ratio of no deflection lengths i.e
The equivalent resistance of the arrangement of resistances shown in adjoining figure between the points A and B is N × 4 ohm then N is:
In the circuit shown here C1 = 6μ F, C2 = 3μ F and battery B = 20V. The switch s1 is first closed. It is then opened and afterwards s2 is closed. What is the charge finally on C2.
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