Electrostatic Potential and Capacitance
- If potential (in volts) in a region is expressed as V(x, y, z) = 6 xy – y + 2yz, the electric field (in N/C) at point (1, 1, 0) is :
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Potential in a region
V = 6xy – y + 2yz As
we know the relation between electric potential and electric field is →E = -dV/dX →E = 
∂V î + ∂V ĵ + ∂V ̂k 
∂x ∂y ∂z
→E = [(6y î + (6x - 1 + 2z)ĵ + (2y)̂k]
→E(1,1,0) = -( 6î + 5ĵ + 2̂k)Correct Option: A
Potential in a region
V = 6xy – y + 2yz As
we know the relation between electric potential and electric field is →E = -dV/dX →E = ∂V î + ∂V ĵ + ∂V ̂k ∂x ∂y ∂z
→E = [(6y î + (6x - 1 + 2z)ĵ + (2y)̂k]
→E(1,1,0) = -( 6î + 5ĵ + 2̂k)
- The diagrams below show regions of equipotentials. [2017]
(a)
(b)
(c)
(d)
A positive charge is moved from A to B in each diagram.
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As the regions are of equipotentials, so Work done W = q∆V
∆V is same in all the cases hence work - done will also be same in all the cases.Correct Option: A
As the regions are of equipotentials, so Work done W = q∆V
∆V is same in all the cases hence work - done will also be same in all the cases.
- A network of four capacitors of capacity equal to C1 = C, C2 = 2C, C3 = 3C and C4 = 4C are conducted to a battery as shown in the figure. The ratio of the charges on C2 and C4 is:
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Equivalent capacitance for three capacitors (C1, C2 & C3) in series is given by 1 = 1 + 1 + 1 = C2C3 + C3C1 + C1C2 Ceq C1 C2 C3 C1C2C3 ⇒ Ceq = C1C2C3 C1C2 + C2C3 + C3C1 ⇒ Ceq = C(2C)(3C) = 6 C C(2C) + (2C)(3C) + (3C)C 11
Charge on capacitors (C1, C2 & C3) in series CeqV = 6C V 11
Charge on capacitor C4 = C4V = 4C V Change on capacitor C2 = (6C/11)V = 6 × 1 = 3 Charge on C4 4CV 11 4 22 Correct Option: B
Equivalent capacitance for three capacitors (C1, C2 & C3) in series is given by 1 = 1 + 1 + 1 = C2C3 + C3C1 + C1C2 Ceq C1 C2 C3 C1C2C3 ⇒ Ceq = C1C2C3 C1C2 + C2C3 + C3C1 ⇒ Ceq = C(2C)(3C) = 6 C C(2C) + (2C)(3C) + (3C)C 11
Charge on capacitors (C1, C2 & C3) in series CeqV = 6C V 11
Charge on capacitor C4 = C4V = 4C V Change on capacitor C2 = (6C/11)V = 6 × 1 = 3 Charge on C4 4CV 11 4 22
- The four capacitors, each of 25µ F are connected as shown in fig. The dc voltmeter reads 200 V. The charge on each plate of capacitor is
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Charge on each plate of each capacitor
Q = ±CV = ± 25 × 10-6 × 200
= ± 5 × × 103 CCorrect Option: B
Charge on each plate of each capacitor
Q = ±CV = ± 25 × 10-6 × 200
= ± 5 × × 103 C
- If the potential of a capacitor having capacity 6 µF is increased from 10 V to 20 V, then increase in its energy will be
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Capacitance of capacitor (C) = 6 µF = 6 ×10-6 F; Initial potential (V1) = 10 V and final potential (V2) = 20 V. The increase in energy (∆U) .
= 1 C(V2² + V1²) 2 = 1 × (6 × 10-6) × [(20)² - (10)²] 2
= (3 × 10-6) × 300 = 9 × 10-4JCorrect Option: C
Capacitance of capacitor (C) = 6 µF = 6 ×10-6 F; Initial potential (V1) = 10 V and final potential (V2) = 20 V. The increase in energy (∆U) .
= 1 C(V2² + V1²) 2 = 1 × (6 × 10-6) × [(20)² - (10)²] 2
= (3 × 10-6) × 300 = 9 × 10-4J
