Electric fields and potential for AP Physics 2

Reading cue. The dashed line runs well beyond both charges on purpose, because not every answer lies between them; the origin sits at +2q, so measure every distance from there.

1. aDetermine Determine the magnitude and direction of the electrostatic force that the +2q charge exerts on the −q charge.

   recall2 marks

   Check answer · 2 marks

   1. F = k(2q)(q)/d² = 2kq²/d², using the separation d marked by the dimension line
   2. the charges are opposite in sign, so the force on −q is attractive and points to the left, towards +2q

2. bDetermine Determine the magnitude and direction of the net electric field at the midpoint of the line joining the two charges.

   routine3 marks

   Check answer · 3 marks

   1. each charge is d/2 from the midpoint; the field of +2q points away from it, to the right, and the field of −q points towards it, also to the right
   2. magnitudes are k(2q)/(d/2)² = 8kq/d² and kq/(d/2)² = 4kq/d²
   3. they add to 12kq/d², directed from +2q towards −q

3. cCalculate Calculate the work an external agent must do to bring a charge +q from far away to the midpoint of the line joining the two charges, taking the potential to be zero at infinity.

   demanding3 marks

   Check answer · 3 marks

   1. potential is a scalar sum: V = k(2q)/(d/2) + k(−q)/(d/2) = 4kq/d − 2kq/d
   2. V = 2kq/d at the midpoint
   3. W = qV = 2kq²/d, and being positive it must be supplied by the external agent

4. dDetermine There is a second point on the dashed line, other than the one between the charges, at which the electric potential is zero. Determine where it lies, and explain why no such point exists to the left of +2q.

   top of the paper4 marks

   Show a hint

   Potential is a scalar, so there is no direction for the two contributions to cancel along — only sign. Write V for a general point in each of the three regions of the dashed line and see which of the three equations has a solution that actually lies in its own region.

   Check answer · 4 marks

   1. for a point at distance x from the origin lying beyond −q, V = 2kq/x − kq/(x − d)
   2. setting V = 0 gives 2(x − d) = x, so x = 2d measured from the origin
   3. that places it a distance d beyond the −q charge, still on the dashed line
   4. to the left of +2q the larger charge is always the nearer one, so 2kq/s always exceeds kq/(s + d) and the sum can never reach zero there

Reading cue. The switch is drawn open, so Q is the quantity held fixed and V is the one free to move; the slab fills the gap completely, so the whole field between the plates is affected.

1. aIndicate Indicate which of the charge on the plates and the potential difference across them is fixed while the slab is inserted, and state the feature of the drawing that settles it.

   recall3 marks

   Check answer · 3 marks

   1. the charge ±Q is fixed
   2. because the switch is drawn open, so no charge can reach or leave the plates
   3. the potential difference is therefore free to change

2. bDetermine The plates have area 2.0 × 10⁻² m² and separation 1.0 mm and carry ±4.0 nC, and the slab has κ = 3.0. Determine the potential difference and the electric field between the plates before and after the slab is inserted. Take ε₀ = 8.85 × 10⁻¹² F m⁻¹.

   routine4 marks

   Check answer · 4 marks

   1. C₀ = ε₀A/d = 1.77 × 10⁻¹⁰ F
   2. V₀ = Q/C₀ = 22.6 V and E₀ = V₀/d = 2.3 × 10⁴ V m⁻¹
   3. C = κC₀ = 5.3 × 10⁻¹⁰ F
   4. V = 7.5 V and E = 7.5 × 10³ V m⁻¹, each reduced by the factor κ

3. cDetermine Determine the energy stored before and after insertion, and hence determine the work done by the person holding the slab as it goes in. State what the sign of that work says about the force on the slab.

   demanding4 marks

   Check answer · 4 marks

   1. U₀ = Q²/2C₀ = 45 nJ
   2. U = U₀/κ = 15 nJ
   3. work done by the person equals ΔU = −30 nJ
   4. negative work means the field pulls the slab in, so the person has to hold it back rather than push it

4. dExplain Explain what would be different if the switch had been left closed, with the battery still connected, while the slab was inserted. Refer to the charge, the stored energy, and the energy supplied by the battery, and support your answer with values.

   top of the paper6 marks

   Check answer · 6 marks

   1. V is now held at 22.6 V, so the charge rises to κQ = 12 nC
   2. stored energy rises to κU₀ = 136 nJ, instead of falling
   3. the battery supplies ΔQ × V = 8.0 nC × 22.6 V = 181 nJ
   4. only 90 nJ of that appears as extra stored energy
   5. the other 90 nJ is the work the field does on the slab, which appears as kinetic energy and is dissipated as the slab is stopped
   6. the slab is pulled inwards either way, so the direction of the force does not depend on the switch

Reading cue. There are no junction dots: one current passes through R and C alike, and with C uncharged at t = 0 the capacitor acts momentarily like a plain wire, putting all of ε across R.

1. aDetermine Determine the current in the circuit immediately after the switch closes, and a long time afterwards.

   recall3 marks

   Check answer · 3 marks

   1. immediately after closing, the uncharged capacitor has no potential difference across it, so all of ε is across R and I₀ = ε/R
   2. after a long time the capacitor potential difference has risen to ε
   3. no potential difference is left across R, so the current is zero

2. bCalculate Take ε = 12 V, R = 47 kΩ and C = 22 μF. Calculate the potential difference across the resistor and across the capacitor, and the current in the circuit, at the moment t = RC.

   routine4 marks

   Check answer · 4 marks

   1. RC = 47 × 10³ × 22 × 10⁻⁶ = 1.03 s
   2. V_C = ε(1 − e⁻¹) = 12 × 0.632 = 7.6 V
   3. V_R = ε − V_C = 4.4 V
   4. I = V_R/R = 9.4 × 10⁻⁵ A

3. cDetermine For the interval from t = 0 to t = RC only, determine the energy delivered by the battery, the energy stored in the capacitor and the energy dissipated in the resistor, and comment on whether the battery's energy divides equally between the two over this interval.

   demanding5 marks

   Check answer · 5 marks

   1. charge delivered q = Cε(1 − e⁻¹) = 1.67 × 10⁻⁴ C
   2. energy from the battery = εq = 2.0 × 10⁻³ J
   3. energy stored = q²/2C = 6.3 × 10⁻⁴ J
   4. energy dissipated = 2.0 × 10⁻³ − 6.3 × 10⁻⁴ = 1.4 × 10⁻³ J
   5. only about 32% is stored over this interval, so the equal split holds for the whole charging process and not for a part of it

4. dDetermine A long time after closing, the switch is reopened. Determine what happens to the charge on the capacitor, describe the smallest change to the circuit that would let the capacitor discharge through R, and determine how long that discharge would take to reduce the charge to 10% of its initial value.

   top of the paper5 marks

   Check answer · 5 marks

   1. the figure is a single loop with the switch in series, so reopening it leaves no conducting path anywhere
   2. the charge Cε = 2.6 × 10⁻⁴ C simply stays on the capacitor, apart from slow leakage
   3. a path is needed that contains C and R but not the open switch, for example a two-way switch that replaces the battery branch with a plain wire
   4. discharge then follows q = Q e^(−t/RC)
   5. t = RC ln10 = 1.03 × 2.30 = 2.4 s