HC Verma Test: Electrostatic Potential & Capacitance - JEE MCQ

The expression of capacitance of metallic sphere is C=4πϵ₀​r (The capacitance of a single spherical conductor means that we imagine a spherical shell of infinite radius surrounding the first conductor.)
Therefore,

 r = C/4πϵ​₀

= 10 ^{− 6} × 9 × 10⁹ = 9 × 10³ m = 9km

The capacity of a parallel plate condenser is given as,
C=εA/​d
When the separation between the plates is halved then the new capacity of the condenser is given as,
C′=εA​/(d/2)​
C′=2C
Thus, the new capacity of parallel plate condenser is 2C.

The energy of a charged capacitor resides in both electric and magnetic fields .
Energy resides in the electric field because of the charges on the capacitor.
Energy resides in the magnetic field because of The Maxwell's displacement current in the capacitor.
 

The capacitance C is the amount of charge stored per volt, or C=QV. The capacitance of a parallel plate capacitor is C=ϵ₀Ad, when the plates are separated by air or free space.

Given,
C₁=0.5 mf
C₂=0.2mf
V=100v
So,
U=(1/2) C_TV²
C_T=C₁C₂/XC₁+C₂=(0.5x0.2/0.5+0.2)mf
C_T=0.14mf
U=(1/2)(0.14mf)(100V)²
=(10.07x1^-6x10⁴)J
=(7x10^-2x10^-6x10⁴)J
=7x10^-4J

First find the charge in 1st capacitor
Q₁= C₁xV₁
=0.003*300
=0.9 C
Similarly, find Q₂
i.e. Q₂=2.5 C
After connecting the capacitors in parallel you get a net capacitance C=8 mF (c₁+c₂).
Thus, final potential is V=Q/C
Q=Q₁+Q₂=3.4 C
V=3.4/0.008
= 425 V
 

Where N is the number of turns of the toroid coil, I is the amount of current flowing and r is the radius of the toroid.
As both capacitors are in parallel so equivalent capacitance C_eq​=C+2C=3C .
Net potential, V_n​=2V−V=V
Energy stored, U=(1/2)​C_eq​V_n²​=(1/2)×3CV²=(3/2)​CV²

Charge on 2μf capacitor=10vx2μF=20μC=Q₁
Charge on 4μf capacitor=4μFx20v=80μC=Q₂
Now, these capacitors are connected in a single loop in given manner,

Now after transfer,

Applying loop rule,
(q/2)+[(60+q)/4]=0=>q=-20μC

Now, drop across the capacitors is,
ΔV_f=40/4v=10v
Final energy stored in capacitor=U_f
= (1/2) (C1+C2)(ΔVf)²=1/2 x6 x100uf=300 μf
Initial energy stored in capacitor=U_i
= (1/2) C₁V₁²+(1/2) C₂V₂²=(1/2)x2x100+ (1/2)x4x400=900 μf
Now, heat evolved in the circuit will be,
U_i-U_f= (900-300) μf=600 μf
Thus, the correct answer would be option B.

If 2μF where not present between B and D, then potential drop across upper 4μF will be less than potential drop across lower 2μF,i.e.,

At equilibrium,
F_electric=F_spring
qE=kx ____(1)
we know that,
C=Q/V
And V=ε.d
C=Aε0/d
Therefore, ε=θ/Aε₀
Here, ε=q/ Aε0
Putting the value in the equation (1)
q/ Aε₀=kx
x=q²/akε₀

⇒q₁​+q₂​+q₃​=0
⇒C₁​(V_o​−V_A​)+C₂​(V_o​−V_B​)+C_3​(V_o​−V_D​)=0
⇒(C_1​+C_2​+C_3​)Vo​=C_1​V_A​+C₂​V_B​+C₃​V_D​
V_o​= C₁​V_A​+C₂​V_B​+C₃​V_D​​/ C₁​+C₂​+C₃​
= [(1×10)+(2×25)+(3×20)​]/(1+2+3)
=(10+50+60)/6
​=20Volts

 

Five capacitors 4μF, 2μF, 4μF, 6μF and 1.2μF are connected as shown in the figure.
Here, Equivalent capacitance = ½,
EMF = 24
Thus, charge passes through each capacitor = 12 μC
Voltage across 4μF = 3V
Voltage across 6μF = 2V
=> Potential difference (V):
V = 3 + 7+ 2 = 12V
Thus, potential difference (V) between the two given points M and N is 12v.
 

a.In this figure, the left and the right branch is symmetry. So the current go to the branch 'ab' for both sides are opposite and equal. Hence it cancels out.
Thus, net charge, Q= 0. ∴V=Q/C​=0/C​=0
∴Vab​=0
b.The net potential, V= Net charge /Net capacitance ​=C1​V1​+C2​V2​+C3​V3/7​​
V=(24+24+24​)/7=72/7​=10.28V 

Given,
Area of plate-S
Initial distance=X₁
Final distance=X₂
Charge given, q
Initial capacitance
C₁=Sε/ X₁
Energy stored
V₁=q²/2C=q²/2(Sε/X₁)
U₁= q²X₁/2Sε
Final capacitance
C₂= Sε/X₂
Energy stored,
U₂= q² x₂/2Sε
Amount of work done=Change in stored energy between capacitor
=U_f-U_i=U₂-U1
=(q²X₂/2Sε)- (q²X₁/2Sε)
Work done=q²/2Sε (X₂-X₁)

When voltage is kept const., the force acing on each plate of capacitor will depend on the distance between the plates. 
From energy conversion,
U_f-U_i=A_cell+A_agent
Or,(1/2)(ε₀S/x2)V2-(1/2)(ε₀S/x1)V²
=[(ε₀S/x₂)- (ε₀S/x1)] V2 +A_agent
(as A_cell=(q_f-q₁)V=(Cf-Ci)V2)
So, A_agent= (ε₀SV²/2)[(1/x₁)-)1/x₂)]

If the charge on left plate of 5μF is −20μC, then the charge on right plate of 5μF is +20μC
So due to the polarization the charge on the left plate of 3μF is negative and the charge on the right plate of 3μF is positive.
thus,  charge on right plate of 3μF is Q=[3/(3+4)]×20=60/7​=8.57μC

The capacitance 4 and (3,5,1) are in parallel so potential across (3,5,1) will be V=24V.
The equivalent capacitance of (3,5,1) is C_eq​=3(5+1)/{3+(5+1)}​=2μF
and Q_eq​=C_eq​V=2×24=48μC
As 3 and (5,1) are in series so charge on 3 and (5,1) is equal to Q_eq​.
Thus potential across (5,1) is V_51​= Q_eq​​/C₅₁​=48/(5+1)​=8V
As 5 and 1 are in parallel so potential difference of capacitors will be same i.e 8V
Now energy stored in 1 is =(1/2)​C₁​V₅₁²​=(1/2)​×10⁻⁶×82=32μJ

Voltage in series connections are,
V₁=C₂/{(C₁+C₂) x V}  V₂={C₁/(C₁+C₂)xV}]
V₁=(1/S) x 200
V₂=(4/S) x200
Ration, V₁/V₂=((1/S) x200)/((4/S)x200)=1:4

Applying Thevenin's theorem we consider the circuit from right side and short circuiting the voltage sources and opening the extreme right resistance we get the equivalent resistance
=1/Rth​​=(1/3R)​+(1/R)​=4/3R=3R/4​ Rth​=Thevenin’s resistance
So now adding the remaining resistance R with Rth​ in series we get the equivalent resistance 
Req​=(3R/4)​+R=7R/4​
And time constant τ=RC
τ=(7/4)​RC