Electrostatic Potential

• E = 500 N C⁻¹ at the point.
• V = 30 V at the same point.
• Both are due to the same point charge.

The magnitude of the charge.

You have two equations and two unknowns (q and r). The elegant move is to divide V by E — the charge and the constant k both cancel, leaving the distance directly. Then substitute back into either equation.

1. Divide: V/E = (kq/r) ÷ (kq/r²) = r.
2. r = 30/500 = 0.06 m.
3. Now use V = kq/r: q = Vr/k = (30 × 0.06)/(9 × 10⁹).
4. = 1.8 / (9 × 10⁹) = 2 × 10⁻¹⁰ C.

• W = 10 J.
• q = 10⁻⁵ C.
• d = 10 cm = 0.1 m, moved against the field.

The field intensity E.

For a uniform field and motion straight along it, work is simply W = qEd. The only place to slip is the unit conversion — 10 cm must become 0.1 m, not 10.

1. Convert: d = 10 cm = 0.1 m.
2. E = W/(qd) = 10 / (10⁻⁵ × 0.1).
3. Denominator: 10⁻⁵ × 10⁻¹ = 10⁻⁶.
4. E = 10 / 10⁻⁶ = 10⁷ N C⁻¹.

• Four charges +Q at the corners of a square of side L.
• All four are the same distance L/√2 from the centre.
• Potential at the centre = V.

The new potential when one charge is reversed to −Q.

Potential is a scalar, so the four contributions simply add with no direction to worry about. Each contributes V/4. Reversing one charge changes its contribution from +V/4 to −V/4 — a change of V/2, not of V/4.

1. All four corners are equidistant from the centre, so each contributes V/4.
2. Reversing one turns its +V/4 into −V/4.
3. New total = 3(V/4) + (−V/4) = 2V/4 = V/2.
4. Equivalently: the change is V/4 − (−V/4) = V/2 subtracted from V, giving V/2.

• An electric dipole.
• Potential measured at distance r, far from the dipole.

How the potential falls with distance.

A dipole is two equal and opposite charges close together, so their potentials very nearly cancel. That partial cancellation costs one extra power of r compared with a single charge: a point charge gives V ∝ 1/r, a dipole gives V ∝ 1/r². The field then falls one power faster still, as 1/r³.

1. For a single point charge, V = kq/r, so V ∝ 1/r.
2. A dipole's two charges are opposite, so their potentials almost cancel, leaving a much smaller residue.
3. That residue is V = kp cosθ/r², so V ∝ 1/r².
4. The field, being the gradient of V, falls one power faster: E ∝ 1/r³.

• Statement-1 about path independence in an electrostatic field.
• Statement-2 about zero work around a closed loop for a conservative force.

The truth of each statement and their relationship.

The electrostatic force is conservative, and that single property gives both statements. Work done depends only on the endpoints, so any path from P to Q gives the same answer; and going out and back around a closed loop returns you to the starting potential, so the total work is zero.

1. The work done by an electrostatic field is W = q(Vₚ − V₪).
2. Only the two potentials appear — the route between them does not enter at all. So Statement-1 is TRUE.
3. For a closed loop the start and end points coincide, so Vₚ = V₪ and W = 0. Statement-2 is TRUE.
4. Both statements are therefore true, whatever view one takes of the explanation link.

• A dipole of moment P in a uniform field E.
• U = −pE cosθ.

The angle at which U is minimum.

The minus sign in U = −pE cosθ does all the work. U is smallest when cosθ is largest, i.e. cosθ = 1 and θ = 0. That is the aligned position — and physically it makes sense: a dipole left alone in a field turns until it lines up, so that must be the position of lowest energy.

1. U = −pE cosθ.
2. For U to be minimum, −cosθ must be minimum, so cosθ must be maximum.
3. cosθ is largest at θ = 0, where cosθ = 1.
4. So U_min = −pE at θ = 0, the aligned position.
5. At θ = π/2, U = 0 — larger than −pE, so not the minimum.

• Dipole starts aligned with the field, θ = 0.
• Rotating to 60° requires work 2 × 10⁻¹⁹ J.
• It is then deflected 30° further, to 90°.

The work done in the second stage.

Work done by an agent rotating a dipole is the change in potential energy, and it depends on the cosines of the two angles, not on the angle swept. So equal angular steps do not require equal work — you must compute each stage from its own endpoints.

1. First stage, 0° to 60°: W₁ = pE(cos0 − cos60) = pE(1 − ½) = ½pE.
2. Given W₁ = 2 × 10⁻¹⁹ J, so pE = 4 × 10⁻¹⁹ J.
3. Second stage, 60° to 90°: W₂ = pE(cos60 − cos90) = pE(½ − 0) = ½pE.
4. So W₂ = ½ × 4 × 10⁻¹⁹ = 2 × 10⁻¹⁹ J — exactly the same as the first stage.

• The definition of an equipotential surface.

Which statement describes it correctly.

An equipotential surface is one on which the potential is the same everywhere. Since work is W = qΔV and ΔV = 0 along the surface, no work is done moving a charge across it. That does not mean the field is zero — it means the field has no component along the surface, which is why field lines always cross an equipotential at right angles.

1. By definition, V is the same at every point on the surface.
2. Work done moving a charge between two points on it: W = q(V₁ − V₂) = q(0) = 0.
3. The field is generally not zero — it is simply perpendicular to the surface, so it does no work along it.
4. Field lines meet an equipotential at 90°, never parallel to it.

• A charged conducting sphere with surface potential 100 V.
• Two statements about the interior.

The truth of each statement and whether one causes the other.

Inside a conductor in electrostatic equilibrium the field must be zero — otherwise free charges would keep moving. And zero field means zero potential gradient, so the potential cannot change from point to point: it is constant, and equal to the surface value. The second fact is a direct consequence of the first.

1. S₁. Free charges rearrange until the interior field is zero. True.
2. S₂. With E = 0 everywhere inside, V has zero gradient, so it cannot vary — it stays at the surface value of 100 V. True.
3. Is S₁ the cause of S₂? Yes: E = −dV/dx, so a zero field is exactly what forces the potential to be constant.
4. So both are true and S₁ causes S₂.

• m = 1 g = 10⁻³ kg, q = 10⁻⁸ C.
• Vₚ = 600 V, V₪ = 0 V.
• v₪ = 20 cm s⁻¹ = 0.2 m s⁻¹.

The velocity at P.

Total energy is conserved: kinetic + electric potential energy is the same at P and at Q. The charge is positive and moves from high potential to low, so it gains kinetic energy on the way — meaning vₚ must be smaller than v₪. That sanity check alone rules out two options.

1. ½mvₚ² = ½mv₪² + q(V₪ − Vₚ) = ½mv₪² − q(600).
2. ½(10⁻³)vₚ² = ½(10⁻³)(0.04) − (10⁻⁸)(600).
3. = 2 × 10⁻⁵ − 6 × 10⁻⁶ = 1.4 × 10⁻⁵.
4. vₚ² = 2 × 1.4 × 10⁻⁵ / 10⁻³ = 0.028.
5. vₚ = √0.028 ≈ 0.167 m s⁻¹, which is indeed less than 0.2. ✓

• Eight drops, each radius r and charge q.
• They merge into a single big drop.

The ratio V_big : V_small.

Two quantities behave differently on merging. Charge simply adds: the big drop carries 8q. But volume adds, not radius — so R³ = 8r³ and R = 2r. The radius only doubles while the charge multiplies by eight, and the potential ratio is 8/2.

1. Charge: Q = 8q.
2. Volume conservation: 8 × (4/3)πr³ = (4/3)πR³, so R³ = 8r³ and R = 2r.
3. V_big = k(8q)/(2r) = 4kq/r.
4. V_small = kq/r.
5. Ratio = 4 : 1.

• q₁ at A and q₂ at B, 30 cm apart.
• q₃ moves along an arc of radius 40 cm centred on A, from C to D.
• C is directly above A; D is on the line AB extended.

The value of k in the expression for ΔU.

The arc is centred on A, so q₃ stays exactly 40 cm from q₁ the whole way — meaning q₁ contributes nothing at all to the change in energy. Only the distance to q₂ changes, and that is the entire calculation.

1. Because the arc is centred on A, the distance A–q₃ is 40 cm at both C and D. q₁ drops out.
2. Distance from q₂ to C: C is 40 cm above A and q₂ is 30 cm along, so √(30² + 40²) = 50 cm.
3. Distance from q₂ to D: D is 40 cm from A along the line, and q₂ is at 30 cm, so 40 − 30 = 10 cm.
4. ΔU = (q₂q₃/4πε₀)(1/0.1 − 1/0.5) = (q₂q₃/4πε₀)(10 − 2).
5. So k = 8q₂.

• Charges of 1 C each at x = 2, 4, 8, 16, … cm.
• All of the same sign.
• Potential wanted at x = 0.

The total potential at the origin.

Potential is a scalar and all the charges have the same sign, so everything simply adds — there is no cancellation. The distances double each time, so the terms 1/r form a geometric progression with ratio ½, which converges to a finite sum. Infinitely many terms does not mean an infinite answer.

1. V = kq(1/0.02 + 1/0.04 + 1/0.08 + …) with all distances in metres.
2. Factor out the first term: = kq × (1/0.02) × (1 + ½ + ¼ + …).
3. The bracket is a GP with first term 1 and ratio ½, so it sums to 1/(1 − ½) = 2.
4. V = 9 × 10⁹ × 1 × 50 × 2 = 9 × 10¹¹ V.