Electrostatic Potential and Capacitance

$F_R=\left\{\frac{kQ{q}_0}{{\left(a-x\right)}^2}-\frac{kQ{q}_0}{{\left(a+x\right)}^2}\right\}$

$=-kQ{q}_0\left\{\frac{a^2+x^2+2ax-a^2-x^2+2ax}{\left(a^2-x^2\right)}\right\}$

$F_R=-\frac{kQ{q}_0\left(4ax\right)}{{\left(a^2-x^2\right)}^2}$

$T=2\pi \sqrt[]{\frac{a^{3}m}{4kQ{q}_0}=\sqrt[]{\frac{4{\pi }^2{a}^{3}m\times 4\pi {\epsilon }_0}{4Q{q}_0}}}=\sqrt[]{\frac{4{\pi }^3{\epsilon }_{0}m{a}^3}{Q{q}_0}}$ .                          

V = $\frac{q_{\left(capacitance\right)}}{C}=\frac{q_1-q_2}{2c}$

For sphere ‘C’ after contacting with ‘A’. q_A = q_C = $\frac{q}{2}.$

offer contacting with ‘B’. q_B = q_C = $\frac{3q}{4}$

$F_{Net}=\left|F_1-F_2\right|$

F₂ = $\frac{K\cdot 3q^2\times 4}{r^{2}8}=\frac{3}{2}\frac{k{q}^2}{r^2}$

$F_1=\frac{9k{q}^2\times 4}{16r^2}=\frac{9k{q}^2}{4r^2}=\frac{9}{4}F$

$\therefore F_{Net}=\left|\frac{9}{4}F-\frac{3}{2}F\right|$

= $\frac{9-6}{4}F=\frac{3}{4}F$

Q = CV

= 50 × 10^-12 × 100

= 5 × 10^-12 × 10³

Q = 5 × 10^-9C

U_i           (Initial Energy)  

=  $\frac{Q^2}{2C}=\frac{25\times 10^{-18}}{2\times 50\times 10^{-12}}$

=  $\frac{1}{4}\times 10^{-6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_1}{C}=\frac{Q_2}{C}=0$    

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2Q_1=Q$     &nb

$F=qE\left[E=\frac{\sigma }{{\epsilon }_0}\right]$

$F=\frac{q\sigma }{{\in }_0}=10N$

Final case – Now one plate is removed, Electric field will becomes half; So, force will become half.

$F\text{'}=\frac{10}{2}=5N$

In forward bias diode act as a short circuit wire. Hence, the equivalent circuit is now.

So,  ${}_{}\frac{}{}$ $V_{ab}=\frac{30}{5+10}*5=10\mathrm{V}$

$\mathrm{}50.00$

Q = CV

$V=\frac{Q}{C}$

Straight line with slope = $\frac{1}{C\text{?}}$ $\frac{}{}$

$Slope=\frac{1}{C}=\frac{1}{2\times 10^{-6}}=5\times 10^5$

Electric Displacement

$\begin{array}{l}\overrightarrow{D}=\in o\overrightarrow{E}\\ \left[D\right]=\left[{\in }_0\overrightarrow{E}\right]=\left[{\in }_0\frac{\sigma }{{\in }_0}\right]\\ \left[D\right]=\left[\sigma \right]\end{array}$

Electric displacement $\left(\overrightarrow{D}\right)$ $\stackrel{}{}$ and surface change density

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{40\times 40k}{40\left(1+k\right)}=24$

$\begin{array}{l}\Rightarrow 40k=24+24k\\ \Rightarrow 16k=24\end{array}$

$k=\frac{24}{16}=\frac{3}{2}=1.5$

K  V  L Loop 1               

6 - l × 1   -l₁ × 1 - 4 = 0

2 = l + l₁