Question
Draw a graph showing the change in potential corresponding to the change in distance from a point charge. For the charged system shown in the figure, prove that the potential difference between points $A$ and $B$ is
✓
Answer
In the figure, the potential at point A due to $+q$ and $-q$ charges respectively is
$V _1=\frac{ l }{4 \pi \epsilon_0} \frac{q}{a}$
And $\quad V _2=\frac{1}{4 \pi \epsilon_0} \frac{(-q)}{(a+d)}$
Therefore the net potential at point A
$V_A=V_1+V_2$
$=\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{a}\right)+\frac{1}{4 \pi \epsilon_0}\left(\frac{-q}{a+d}\right)$
$V _{ A }=\frac{1}{4 \pi \epsilon_0}\left(\frac{1}{a}-\frac{1}{a+d}\right)$
$V _{ A }=\frac{q}{4 \pi \epsilon_0}\left(\frac{a+d-a}{a(a+d)}\right)=\frac{1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}$
Similarly net potential at point B
$V _{ B }=\frac{q}{4 \pi \epsilon_0}\left(\frac{1}{a+d}-\frac{1}{a}\right)$
$=\frac{-1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}$
Therefore,
$V_A-V_B$$\frac{1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}-\left(\frac{-1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}\right)$
$\begin{aligned} V _{ A }- V _{ B } & =\frac{1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}[1+1] \\ & =\frac{2 q d}{4 \pi \epsilon_0 a(a+d)}\end{aligned}$
$V _{ A }- V _{ B }=\frac{q}{2 \pi \epsilon_0 a} \cdot \frac{d}{a+d}$
$V _1=\frac{ l }{4 \pi \epsilon_0} \frac{q}{a}$
And $\quad V _2=\frac{1}{4 \pi \epsilon_0} \frac{(-q)}{(a+d)}$
Therefore the net potential at point A
$V_A=V_1+V_2$
$=\frac{1}{4 \pi \epsilon_0}\left(\frac{q}{a}\right)+\frac{1}{4 \pi \epsilon_0}\left(\frac{-q}{a+d}\right)$
$V _{ A }=\frac{1}{4 \pi \epsilon_0}\left(\frac{1}{a}-\frac{1}{a+d}\right)$
$V _{ A }=\frac{q}{4 \pi \epsilon_0}\left(\frac{a+d-a}{a(a+d)}\right)=\frac{1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}$
Similarly net potential at point B
$V _{ B }=\frac{q}{4 \pi \epsilon_0}\left(\frac{1}{a+d}-\frac{1}{a}\right)$
$=\frac{-1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}$
Therefore,
$V_A-V_B$$\frac{1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}-\left(\frac{-1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}\right)$
$\begin{aligned} V _{ A }- V _{ B } & =\frac{1}{4 \pi \epsilon_0} \frac{q d}{a(a+d)}[1+1] \\ & =\frac{2 q d}{4 \pi \epsilon_0 a(a+d)}\end{aligned}$
$V _{ A }- V _{ B }=\frac{q}{2 \pi \epsilon_0 a} \cdot \frac{d}{a+d}$
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A plane electromagnetic wave travels in free space along $x-$axis. At a particular point in space, the electric field along $y$-axis is $9.3 V m ^{-1}$. The magnetic induction $(B)$ along $z -$axis is
$(a) \ 3.1 \times 10^{-8} T$
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$(iv)$ A plane electromagnetic wave travelling along the $x-$direction has a wavelength of $3 \ mm$ . The variation in the electric field occurs in the $y-$direction with an amptitude $66 V m ^{-1}$. The equations for the electric and magnetic fields as a function of $x$ and $t$ are respectively
$ \text { a) } E_y =11 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right), \text { b) } E_y =66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right),$
$B_y =11 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right) B_z =2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)$
$\text { c) } E_x =33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right), \text { d) } E_y =33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right),$
$B_x =11 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right) B_z =1.1 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)$
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$(a) \vec{E}= B _0 c \sin ( kx +\omega t ) \hat{k} V / m$
$(b) \vec{E}=- B _0 c \sin ( kx -\omega t ) \hat{k} V / m$
$(c) \vec{E}=- B _0 c \sin ( kx +\omega t ) \hat{k} V / m$
$(d) \vec{E}=\frac{B_0}{c} \sin ( kx +\omega t ) \hat{k} V / m$
$(ii)$ The electric field component of a monochromatic radiation is given by $\vec{E}=2 E _0 \hat{i} \cos kz \cos \omega t$. Its magnetic field $\vec{B}$ is then given by
$(a) -\frac{2 E_0}{c} \hat{j} \sin kz \sin \omega t$
$(b) \frac{2 E_0}{c} \hat{j} \sin kz \sin \omega t$
$(c) \frac{2 E_0}{c} \hat{j} \sin kz \cos \omega t$
$(d) \frac{2 E_0}{c} \hat{j} \cos kz \cos \omega t$
$(iii)$ A plane em wave of frequency $25 \ MHz$ travels in a free space along $x -$direction. At a particular point in space and time, $E =(6.3 \hat{j}) V / m$. What is magnetic field at that time?
$(a) \ 0.089 \mu T$
$(b) \ 0.124 \mu T$
$(c) \ 0.021 \mu T$
$(d) \ 0.095 \mu T$
OR
A plane electromagnetic wave travels in free space along $x-$axis. At a particular point in space, the electric field along $y$-axis is $9.3 V m ^{-1}$. The magnetic induction $(B)$ along $z -$axis is
$(a) \ 3.1 \times 10^{-8} T$
$(b) \ 3 \times 10^{-5} T$
$(c) \ 3 \times 10^{-6} T$
$(d) \ 9.3 \times 10^{-6} T$
$(iv)$ A plane electromagnetic wave travelling along the $x-$direction has a wavelength of $3 \ mm$ . The variation in the electric field occurs in the $y-$direction with an amptitude $66 V m ^{-1}$. The equations for the electric and magnetic fields as a function of $x$ and $t$ are respectively
$ \text { a) } E_y =11 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right), \text { b) } E_y =66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right),$
$B_y =11 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right) B_z =2.2 \times 10^{-7} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)$
$\text { c) } E_x =33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right), \text { d) } E_y =33 \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right),$
$B_x =11 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right) B_z =1.1 \times 10^{-7} \cos \pi \times 10^{11}\left(t-\frac{x}{c}\right)$