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 120V. It's energy is 1×10−5J. If battery is there and the space between plates is filled up with a dielectric medium (εr=5). Its new energy is
The area of the positive plate is 125cm2 and the area of the negative plate is 100cm2, They are parallel to each other and are separated by 0.5 cm . The capacity of a condenser with air as dielectric is :
(ε0=8.9×10−12C2N−1M−2)
Two identical metal plates, separated by a distance d form a parallel plate capacitor. A metal sheet of thicknessd2 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 4ϵ0Ad and 4ϵ0Ad
⇒1Cq=d4ϵ0A+d4ϵ0A
⇒Ceff=2ϵ0Ad
C2C1=2ϵ0Adϵ0Ad=2:1
When a dielectric slab of thickness 4cm is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by 3cm to restore the capacity to it's original value. The dielectric constant of the slab is
Electric potential, V=14πϵ0qr=14πϵ0q√(4−1)2+(7−3)2+(2−2)2=10−84π×8.854×10−12×5=18volt
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. Q1 and Q2 are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are q1 and q2 . Then :
Two capacitors of capacities 3μF and 6μF are connected in series and connected to 120V. The potential differences across 3μF is V0 and the charge here is q0. We have :
A)q0=40μC B) V0=60V
C)V0=80V D)q0=240μC
A potential difference of 300 volts is applied to a combination of 2.0μF and 8.0μF capacitors connected in series. The charge on the 2.0μF capacitor is :
The capacitors of three capacities are in the ratio 1 : 2 :Their equivalent capacity when connected in parallel is 6011μF more than that when connected in series. The individual capacities are :
A parallel plate capacitor with plates separated by air acquires 1 μC 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 14th 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 Q0,V0,E0 and U0 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 C1(fig. a). A dielectric of dielectric constant ‘K’ is inserted as shown in figure ‘b’ and ‘c’. If C2 and C3 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=4kq2√x2+a2/2
˙U=−8kqx22√x2+a2/23
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 WB=k(−e)(−q)rb=kqerb where rb be the distance between −e and −q.
similarly, WC=k−qerc and WD=k−qerd and WA=0
Thus, the highest potential energy point will be B.
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