Two equal point charges are fixed at $$x = -a$$ and $$x= +a$$ on the x- axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to :
Explanation
A parallel plate condenser has initially air medium between the plates. If a slab of dielectric constant 5 having thickness half the difference of separation between the plates is introduced, the percentage increase in its capacity is :
Three capacitors $$3 \mu F, 10 \mu F$$ and $$15 \mu F$$ are connected in series to a voltage source of 100V. The charge on $$15 \mu F$$ is :
A condenser is charged to a potential difference of $$120 \; V$$. It's energy is $$1 \times 10^{-5}\; J$$. If battery is there and the space between plates is filled up with a dielectric medium $$\left ( \varepsilon _{r}=5\right )$$. Its new energy is
The area of the positive plate is $$125\;cm^{2}$$ and the area of the negative plate is $$100\;cm^{2}$$, They are parallel to each other and are separated by 0.5 cm . The capacity of a condenser with air as dielectric is :
$$\left ( \varepsilon _{0}=8.9\times 10^{-12}\;C^{2}N^{-1}M^{-2} \right )$$
Two identical metal plates, separated by a distance d form a parallel plate capacitor. A metal sheet of thickness$$\dfrac{d}{2} $$ of the same area as that of either plate, is inserted between the plates. The ratio of the capacitance's after the insertion of the sheet to that before insertion is:
The effective capacitance of above capacitance is series combination of $$\displaystyle \frac{ {4\epsilon}_{0} A}{d} $$ and $$ \dfrac{{4\epsilon}_{0}A }{d} $$
$$\Rightarrow \dfrac{1}{C_{q}}=\dfrac{d}{4 {\epsilon}_{0} A}+\dfrac{d}{4 {\epsilon}_{0}A} $$
$$\Rightarrow C_{eff}=\dfrac{2 {\epsilon}_{0}A}{d} $$
$$\dfrac{C_{2}}{C_{1}}=\dfrac{\dfrac{2 {\epsilon}_{0}A}{d}}{\dfrac{{\epsilon}_{0}A}{d}}=2:1$$
When a dielectric slab of thickness $$4 \;cm$$ is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by $$3 \; cm$$ to restore the capacity to it's original value. The dielectric constant of the slab is
Electric potential$$,\ V=\dfrac{1}{4\pi \epsilon_0}\dfrac{q}{r}=\dfrac{1}{4\pi \epsilon_0}\dfrac{q}{\sqrt{(4-1)^2+(7-3)^2+(2-2)^2}}=\dfrac{10^{-8}}{4\pi \times 8.854\times 10^{-12}\times 5}=18 volt$$
Two identical capacitors 1 and 2 are connected in series to a battery as shown in figure. Capacitor 2 contains a dielectric slab of dielectric constant K as shown. $$Q_{1}$$ and $$Q_{2}$$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $$q_{1}$$ and $$q_{2}$$ . Then :
Two capacitors of capacities 3$$\mu $$F and 6$$\mu F$$ are connected in series and connected to 120V. The potential differences across 3$$\mu $$F is $$V_{0}$$ and the charge here is $$q_{0}$$. We have :
A)$$q_{0}=40\mu C$$ B) $$V_{0}=60V$$
C)$$V_{0}=80V$$ D)$$q_{0}=240\mu C$$
A potential difference of $$300$$ volts is applied to a combination of $$2.0\mu $$F and $$8.0\mu $$F capacitors connected in series. The charge on the $$2.0\mu $$F capacitor is :
The capacitors of three capacities are in the ratio 1 : 2 :Their equivalent capacity when connected in parallel is $$\frac{60}{11}\mu F$$ more than that when connected in series. The individual capacities are :
A parallel plate capacitor with plates separated by air acquires 1 $$\mu $$C of charge when connected to a battery of 500V. The plates still connected to the battery are then immersed in benzene ($$k=2.25$$). Then a charge flows from the battery is :
A capacitor of capacitance $$C$$ has charge $$Q$$ and stored energy $$W$$. If the charge is increased to $$2Q$$, the stored energy would be:
The distance between the plates of a condenser is reduced to $$\frac{1}{4}th$$ and the space between the plates is filled up by a medium of dielectric constant K(2.8). The capacity is increased by :
The capacity of a parallel plate condenser with air medium is C . If half of the space between the plates is filled with a slab of dielectric constant K as shown in the figure, then the capacity becomes :
A parallel plate capacitor of capacitance C is charged and disconnected from the battery. The energy stored in it is E. If a dielectric slab of dielectric constant 6 is inserted between the plates of the capacitor then energy and capacitance will become :
A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric filed and energy associated with this capacitor are gives by $$Q_{0},V_{0},E_{0}$$ and $$U_{0}$$ respectively. A dielectric slab is now introduced to fill the space between the plates with the battery still in connection. The corresponding quantities are now given by $$Q$$, $$V$$, $$E$$ and $$U$$ are related to the previous one as:
The capacitance of a parallel plate condenser is $$C_{1}$$(fig. a). A dielectric of dielectric constant ‘K’ is inserted as shown in figure ‘b’ and ‘c’. If $$C_{2}$$ and $$C_{3}$$ are the capacitances in figures ‘b’ and ‘c’ then :
Now potential energy of charge +q at any distance $$x$$ from origin along x-axis is given
$$U=\dfrac { 4k{ q }^{ 2 } }{ \sqrt { { x }^{ 2 }+{ { a }^{ 2 } }/{ 2 } } } $$
$$\dot { U } =-\dfrac { 8k{ qx }^{ 2 } }{ 2{ \sqrt { { x }^{ 2 }+{ { a }^{ 2 } }/{ 2 } } }^{ 3 } } $$
Now as we can see for $$x<0$$ the slope potential energy curve is positive and at
$$x=0$$ the slope is zero and at
$$x>0$$ the slope is negative
Hence the correct answer is option (B)
Potential energy when electron at B is $$\displaystyle W_B=k\dfrac{(-e)(-q)}{r_b}=k\dfrac{qe}{r_b}$$ where $$r_b$$ be the distance between $$-e$$ and $$-q$$.
similarly, $$\displaystyle W_C=k\dfrac{-qe}{r_c} $$ and $$\displaystyle W_D=k\dfrac{-qe}{r_d} $$ and $$W_A=0$$
Thus, the highest potential energy point will be B.
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