Physics Electrostatic Potential and Capacitance

V = E (1 - e? /τ)
Where τ = RC = 100 * 10? = 10? sec
50V = 100V (1 - e? /10? )
1/2 = 1 - e? ¹?
1/2 = e? ¹?
-ln2 = -10? t
t = ln2/10?
t = 0.693 * 10? sec

dC? = (ε? + kx)A / dx [For 0 < x < d/2]
1/C? = ∫ dx / (ε? + kx)A) from 0 to d/2
= (1/Ak) [ln (ε? + kx)] from 0 to d/2
= (1/kA) ln (1 + kd/ (2ε? )
C? = kA / ln (1 + kd/ (2ε? )

Similarly dC? = (ε? + k (d-x)A / dx [For d/2 ≤ x ≤ d]
C? = kA / ln (1 + kd/ (2ε? )
Clearly, C? = C? = C
For series combination:
C_eq = C? / (C? + C? ) = C/2 = kA / (2ln (2ε? + kd)/2ε? )

Enet = Eo/k
Enet = E_free - E_bound = qf/Aε? - qb/Aε?
Eo = qf/Aε?
So, (qf-qb)/Aε? = qf/ (kAε? )
qf - qb = qf/k
qb = qf (1 - 1/k)

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(-2Q\right)}{{\left(\frac{\sqrt[]{3}}{2}a\right)}^2}.\frac{1}{\sqrt[]{3}}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)=\frac{|-2Q}{3\sqrt[]{3}\pi {\epsilon }_0{a}^2}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)$

When connected in series, equivalent capacitance,

$C_1=\frac{C*C}{C+C}=\frac{C}{2}$            

When connected in parallel, equivalent capacitance

C₂ = C + C = 2C

$\frac{C_1}{C_2}=\frac{C/2}{2C}=\frac{1}{4}$            

$\lambda =\frac{c}{v}=\frac{Q^3}{W^2}=\frac{I^3{T}^3}{M^2{L}^4{T}^{-4}}\Rightarrow \left[\lambda \right]=\left[M^{-2}{L}^{-4}{T}^7{l}^3\right]$                                                                             

According question, we can write

$\frac{C_1*C_2}{C_1+C_2}=\frac{15}{4}\left(C_1+C_2\right)\Rightarrow 4C_1{C}_2=15\left(C_1^2+C_2^2+2C_1{C}_2\right)$

$\Rightarrow 4\left(\frac{C_2}{C_1}\right)=15{\left(\frac{C_2}{C_1}\right)}^2+30\left(\frac{C_2}{C_1}\right)+15\Rightarrow 15x^2+26x+15=0,$ where x = $\frac{C_2}{C_1}$

$\Rightarrow x=\frac{-26\pm \sqrt[]{676-900}}{30}$

Since x cannot be real, so this Question has been dropped by NTA

After contact, charges an each sphere will be

$\frac{q_1+q_2}{2}=1nc\Rightarrow F=\frac{k{q}_1{q}_2}{r^2}=36*10^{-9}N$

$Tsin\theta =\frac{k{q}^2}{r^2}$

$Tcos\theta =mg$

$tan\theta =\frac{k{q}^2}{mg{r}^2}$

Using the above equation find the value of 'q'

Given  $V^{\prime }=V=$ Constant

 (i) $C^{\prime }=\frac{{\epsilon }_{0}A}{d^{\prime }},C=\frac{{\epsilon }_{0}A}{d}$

$d^{\prime }<d$

$C^{\prime }>C$
Hence, final capacitance greater than initial capacitance,

(ii)  $U^{\prime }=\frac{1}{2}{C}^{\prime }{V}^2$

$U=\frac{1}{2}C{V}^2$

$U^{\prime }>U$

Hence final energy is greater than initial energy

(iii)  $\frac{Q^{\prime }}{V^{\prime }}=C^{\prime }$ and $\frac{Q}{V}=C$

$\frac{Q^{\prime }}{V^{\prime }}\ne \frac{Q}{V}$

(iv) Product of charge and voltage

$X^{\prime }=Q^{\prime }V=C^{\prime }{V}^2$

$X=QV=C{V}^2$

$X^{\prime }>X$