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 3μF,10μF and 15μF are connected in series to a voltage source of 100V. The charge on 15μF 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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