Questions: Electrostatic Potential and Capacitance

Q1. Define electric potential at a point and write its SI unit. Why is it a scalar even though the field is a vector?

Answer (English): The potential at a point is the work per unit positive charge that an external agent does in carrying a small test charge from infinity to that point, at negligible speed so its kinetic energy does not change. The SI unit is the volt, which is one joule per coulomb. Potential is a scalar because it is built from work, and work is a scalar. Add potentials of several charges as ordinary numbers, while fields must be added as vectors.

Hinglish: Electric potential woh kaam hai jo ek unit positive test charge ko infinity se us point tak laane mein kiya jata hai, bilkul dheere, bina kinetic energy badle. Iski SI unit volt hai, matlab joule per coulomb. Potential scalar hai kyunki work scalar hota hai. Isliye kai charges ke potentials seedha add ho jate hain, jabki fields ko vector ki tarah add karna padta hai.

Q2. Derive the relation between electric field and potential.

Answer (English): Take two points separated by a small displacement dl along the field. The work done by the field on a unit charge in that displacement is E dl, and this equals the fall in potential, so E dl = -dV. Therefore E = -dV/dl, which in one dimension reads E = -dV/dx. The negative sign says the field points from higher potential to lower potential. The same relation in three dimensions uses the gradient, and it shows that a region of constant potential has no field.

Hinglish: Field ke along chhota displacement dl lo. Field jo kaam karti hai unit charge par woh E dl hai, aur yeh potential ke girne ke barabar hai, isliye E dl = -dV aur E = -dV/dl milta hai. Negative sign ka matlab field higher potential se lower potential ki taraf hoti hai. Jahan potential constant hota hai wahan field zero hoti hai, aur yahi equipotential ka matlab hai.

Q3. What are equipotential surfaces? Why is the electric field always perpendicular to them?

Answer (English): An equipotential surface is a surface on which every point has the same potential. Moving a charge along such a surface needs no work, so the field must have no component along the surface. If the field had a tangential component, work would be done in a displacement along the surface, and the potential would change, which contradicts the definition. So the field meets the equipotential surface at a right angle everywhere. Equipotential surfaces and field lines cross at right angles, and two different equipotential surfaces never meet.

Hinglish: Equipotential surface par har point ka potential same hota hai. Is surface ke along charge move karne mein work zero chahiye, isliye field ka surface ke along koi component nahi ho sakta. Agar component hota to displacement par work hota aur potential badalta, jo definition ke against hai. Isliye field har point par surface ke perpendicular hoti hai. Equipotential surfaces field lines ko right angle par kaatti hain aur kabhi intersect nahi karti.

Q4. Two equal positive charges are placed a distance apart. Find the potential and the field at the midpoint.

Answer (English): At the midpoint each charge is at the same distance d from the point, so the potentials add: V = 2 k q / d. The potential is positive and not zero. The two fields, however, are equal in size and opposite in direction, so their vector sum is zero. This is a useful example of the difference between the two quantities: the field can vanish at a point where the potential is large.

Hinglish: Midpoint par dono charges ki distance same d hoti hai, isliye potentials add ho jate hain: V = 2 k q / d, jo positive hai. Lekin dono charges ki fields size mein barabar aur direction mein opposite hoti hain, isliye net field zero ho jata hai. Isse clear hota hai ki potential zero na ho tab bhi field zero ho sakti hai, aur ulta bhi.

Q5. Derive the capacitance of a parallel plate capacitor with a dielectric between the plates.

Answer (English): Let the plates have area A and separation d. Without a dielectric the field between the plates is E0 = Q / (epsilon0 A) and the potential difference is E0 d, so C0 = epsilon0 A / d. With a dielectric of constant K the induced surface charges reduce the field to E0 / K. The potential difference falls to E0 d / K, while the charge stays the same. Dividing charge by the new potential difference gives C = K epsilon0 A / d, so the capacitance is multiplied by K.

Hinglish: Plate area A aur separation d lo. Bina dielectric ke field E nought = Q / (epsilon nought A) aur voltage E nought d, isliye C nought = epsilon nought A / d. Dielectric daalne par induced charges field ko E nought / K kar dete hain aur voltage bhi K guna kam ho jata hai, jabki charge same rehta hai. Isliye nayi capacitance K epsilon nought A / d ho jati hai, matlab capacitance K guna badh jati hai.

Q6. Explain why the electric field inside a conductor is zero and why the charge sits on the surface.

Answer (English): In equilibrium the free charges in a conductor have stopped moving. A free charge stops only when the force on it is zero, and the force is qE, so E must be zero everywhere inside the conductor. Now take a Gaussian surface just inside the surface of the conductor. The field is zero on this surface, so the flux is zero, and by Gauss's law the enclosed charge is zero. Therefore no net charge sits in the interior, and any excess charge must be on the outer surface. The surface is an equipotential, and the field just outside is perpendicular to it.

Hinglish: Equilibrium mein conductor ke free charges ruk jate hain, aur charge par force zero hoti hai tabhi rukta hai, isliye andar field zero honi chahiye. Conductor ki surface ke thoda andar ek Gaussian surface lo: us par field zero hai, isliye flux zero, aur Gauss's law se andar ka charge zero. Iska matlab interior mein net charge nahi hota, aur extra charge outer surface par baith jata hai. Surface equipotential hoti hai aur bahar ki field uske perpendicular hoti hai.

Q7. A capacitor is charged by a battery and then the battery is disconnected. A dielectric slab is inserted. What happens to the charge, potential difference, field and energy?

Answer (English): With the battery disconnected the charge on the plates cannot change. The capacitance rises to K times its original value. The potential difference falls to V / K because V = Q / C. The field between the plates falls to E / K. The stored energy falls to U / K, because U = Q^2 / 2C and C has increased. This is the opposite of the case where the battery stays connected, and the difference comes entirely from which quantity is held fixed.

Hinglish: Battery hata dene par plate ka charge fixed rehta hai. Capacitance K guna badh jati hai. Voltage V = Q / C se K guna kam ho jata hai, field E / K ho jati hai aur energy Q^2 / 2C se K guna kam ho jati hai. Battery lagi hone par iska ulta hota hai, kyunki tab voltage fixed rehta hai. Isliye sawaal mein pehle decide karo battery lagi hai ya nahi.

Q8. Three capacitors of 2, 3 and 6 microfarad are joined in series. Find the equivalent capacitance.

Answer (English): For a series combination the reciprocals add: 1/C = 1/2 + 1/3 + 1/6, in units of 1/microfarad. Convert to sixths: 3/6 + 2/6 + 1/6 = 6/6 = 1. So 1/C = 1 per microfarad, which gives C = 1 microfarad. The equivalent capacitance is smaller than the smallest capacitor, as it must be for a series combination.

Hinglish: Series combination mein reciprocal add hote hain: 1/C = 1/2 + 1/3 + 1/6. Isko chhah ke bhinno mein likho: 3/6 + 2/6 + 1/6 = 1, isliye 1/C = 1 aur C = 1 microfarad. Series mein equivalent capacitance sabse chhoti capacitor se bhi chhoti hoti hai, yeh check karne ka aasan tareeka hai.

Q9. Derive the expression for the energy stored in a capacitor.

Answer (English): At an intermediate stage of charging, let the charge on the capacitor be q and the potential difference be v = q / C. Work done in adding a small charge dq against the potential is dW = v dq = (q / C) dq. Integrate from zero to the final charge Q: W = Q^2 / 2C. This work is stored as the energy of the capacitor. Using Q = C V gives U = (1/2) C V^2, and using V = Q / C gives U = Q V / 2. All three forms are equivalent.

Hinglish: Charging ke beech mein charge q ho to voltage v = q / C hota hai. Chhota charge dq add karne mein work dW = v dq = (q / C) dq. Zero se final charge Q tak integrate karo: W = Q^2 / 2C. Yehi energy store hoti hai. Q = C V rakhne par U = 1/2 C V^2 aur V = Q / C rakhne par U = Q V / 2 milta hai. Teeno forms barabar hain, jo sawaal mein aasani ho woh use karo.

Q10. Why does the capacitance of a parallel plate capacitor increase when a dielectric is inserted?

Answer (English): The dielectric polarises in the field between the plates. Its induced surface charges set up a field that opposes the original field, so the net field between the plates is weaker for the same charge on the plates. A weaker field means a smaller potential difference across the same separation. Capacitance is charge divided by potential difference, so a smaller potential difference for the same charge gives a larger capacitance. The factor by which the field is weakened is the dielectric constant K, so the capacitance becomes K times larger.

Hinglish: Dielectric field mein polarise ho jata hai. Uski induced surface charges ek ulti field banati hain jo original field ko kam karti hai, isliye same charge par andar ki net field kamzor ho jati hai. Kamzor field ka matlab same separation par kam voltage. Capacitance Q / V hai, isliye same charge par kam voltage dene se capacitance badh jati hai. Jitna field kam hota hai utna hi K ka factor hota hai, isliye capacitance K guna badh jati hai.

Quiz below: five questions on potential, equipotential surfaces, capacitance combinations, dielectric changes and stored energy. Check each wrong answer against the formula sheet and the capacitance page.