The following question which appeared in IIT 1997 Entrance test paper checks whether you have a thorough understanding of the basic relation between the electric field and potential:
A non-conducting ring of radius 0.5m carries a total charge of 1.11×10–10 C distributed non uniformly on its circumference producing an electric field E everywhere in space. The value of the line integral ∫ –E.dl between the limits l = ∞ to l = 0 ( l = 0 being the centre of the ring) in volts is
(a) +2 (b) –1 (c) –8 (d) zero
The line integral ∫ –E.dl between the limits l = ∞ and l = 0 gives the work done in bringing unit positive charge from infinity to the centre of the ring and therefore is equal to the electric potential at the centre of the ring. If the total charge on the ring is Q coulomb, the potential at the centre is V = (1/4πε0)×Q/r where ‘r’ is the radius of the ring. Therefore, V = 9×109 ×1.11×10–10/0.5 = 2, very nearly. So, the correct option is (a).
Now, consider the following MCQ:
A dielectric slab (dielectric constant = K) of thickness ‘t’ is placed between the plates of a parallel plate air capacitor. If the capacitance of the capacitor is to be restored to the original value, the separation between the plates is to be increased by
(a) kt (b) k/t (c) t/k (d) t –(k/t) (e) t –(t/k)
When a dielectric slab of thickness ’t’ is introduced between the plates, the electric fields in the air space and in the dielectric space are respectively q/ε0A and q/Kε0A where ‘q’ is the charge on each plate (+q on one, –q on the other) and A is the area of the plate. The P.D. between the plates is V = (q/ε0A)(d–t)+(q/Kε0A)t = (q/ε0A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is C = q/V = ε0A/[d–t+(t/K)] = ε0A/[d–(t – t/K)].
Since the capacitance of the capacitor with air filling the entire space between the plates is ε0A/d, the effect of introducing the dielectric slab of thickness ‘t’ is to reduce the thickness of air by t– (t/K). In order to restore the capacitance to the original value, the separation between the plates is to be increased by t– (t/K).
The following MCQ appeared in Kerala Engineering Entrance 2003 question paper:
A parallel plate capacitor has a capacitance of 100 pF when the plates of the capacitor are separated by a distance of ‘t’. Then a metallic foil of thickness t/3 is introduced between the plates. The capacitance will then become
(a) 100 pF (b) (3/2)100 pF (c) (2/3)100 pF (d) (1/3)100 pF (e) (1/2)100 pF
As shown in the previous discussion, the capacitance of a parallel plate capacitor with a dielectric slab of thickness ‘t’ between the plates separated by a distance ‘d’ is given by,C = ε0A/[d–t+(t/K)]. In the present problem, ‘d’ is to be replaced by ‘t’ and ‘t’ is to be replaced by t/3. Further, the dielectric constant K is to be replaced by ∞ since the dielectric constant of a conductor is infinite. The capacitance therefore becomes ε0A/(t–t/3) = (3/2) ε0A/t = (3/2)100 pF (since the original capacitance with air alone as the dielectric is ε0A/t = 100 pF).
A non-conducting ring of radius 0.5m carries a total charge of 1.11×10–10 C distributed non uniformly on its circumference producing an electric field E everywhere in space. The value of the line integral ∫ –E.dl between the limits l = ∞ to l = 0 ( l = 0 being the centre of the ring) in volts is
(a) +2 (b) –1 (c) –8 (d) zero
The line integral ∫ –E.dl between the limits l = ∞ and l = 0 gives the work done in bringing unit positive charge from infinity to the centre of the ring and therefore is equal to the electric potential at the centre of the ring. If the total charge on the ring is Q coulomb, the potential at the centre is V = (1/4πε0)×Q/r where ‘r’ is the radius of the ring. Therefore, V = 9×109 ×1.11×10–10/0.5 = 2, very nearly. So, the correct option is (a).
Now, consider the following MCQ:
A dielectric slab (dielectric constant = K) of thickness ‘t’ is placed between the plates of a parallel plate air capacitor. If the capacitance of the capacitor is to be restored to the original value, the separation between the plates is to be increased by
(a) kt (b) k/t (c) t/k (d) t –(k/t) (e) t –(t/k)
When a dielectric slab of thickness ’t’ is introduced between the plates, the electric fields in the air space and in the dielectric space are respectively q/ε0A and q/Kε0A where ‘q’ is the charge on each plate (+q on one, –q on the other) and A is the area of the plate. The P.D. between the plates is V = (q/ε0A)(d–t)+(q/Kε0A)t = (q/ε0A)[d–t+(t/K)]. The capacitance of the system on introducing the dielectric is C = q/V = ε0A/[d–t+(t/K)] = ε0A/[d–(t – t/K)].
Since the capacitance of the capacitor with air filling the entire space between the plates is ε0A/d, the effect of introducing the dielectric slab of thickness ‘t’ is to reduce the thickness of air by t– (t/K). In order to restore the capacitance to the original value, the separation between the plates is to be increased by t– (t/K).
The following MCQ appeared in Kerala Engineering Entrance 2003 question paper:
A parallel plate capacitor has a capacitance of 100 pF when the plates of the capacitor are separated by a distance of ‘t’. Then a metallic foil of thickness t/3 is introduced between the plates. The capacitance will then become
(a) 100 pF (b) (3/2)100 pF (c) (2/3)100 pF (d) (1/3)100 pF (e) (1/2)100 pF
As shown in the previous discussion, the capacitance of a parallel plate capacitor with a dielectric slab of thickness ‘t’ between the plates separated by a distance ‘d’ is given by,C = ε0A/[d–t+(t/K)]. In the present problem, ‘d’ is to be replaced by ‘t’ and ‘t’ is to be replaced by t/3. Further, the dielectric constant K is to be replaced by ∞ since the dielectric constant of a conductor is infinite. The capacitance therefore becomes ε0A/(t–t/3) = (3/2) ε0A/t = (3/2)100 pF (since the original capacitance with air alone as the dielectric is ε0A/t = 100 pF).
You will find more multiple choice questions with solution at physicsplus: Questions on Electrostatics
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