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Learn/ 12th Physics/ Unit 12 /Numericals

Unit 12: Electrostatics — Numericals

12th Class Physics · Unit 12: Electrostatics

12.1.Charges q₁ = 100 μC and q₂ = 50 μC are located in xy-plane at positions r₁ = 3.0 j and r₂ = 4.0 i respectively, where the distances are measured in metres. Calculate the force on q₂.
Given
First charge q1 = 100 μC = 100 × 10-6 C
Second charge q2 = 50 μC = 50 × 10-6 C
Position vector vec{r}21 = vec{r}2 - vec{r}1 = 4hat{i} - 3hat{j}
Formula
Coulomb's law (vector form) vec{F}21 = Kq1 q2r2hat{r}21
Unit vector hat{r}21 = frac{4hat{i} - 3hat{j}{sqrt{42 + 32} = frac{4hat{i} - 3hat{j}{5}
Working
Substituting values vec{F}21 = 9 × 109 × frac{100 × 10-6 × 50 × 10-6{25} × frac{4hat{i} - 3hat{j}{5}
Components = 1.44hat{i} - 1.08hat{j}
Result
Magnitude of force F21 = sqrt{(1.44)2 + (1.08)2} = 1.8 N
12.2.Two positive point charges q₁ = 16.0 μC and q₂ = 4.0 μC are separated by a distance of 3.0 m, as shown in figure. Find the spot on the line joining the two charges where electric field is zero.
Given
First charge q1 = 16.0 μC = 16.0 × 10-6 C
Second charge q2 = 4.0 μC = 4.0 × 10-6 C
Distance between charges r = 3.0 m
Formula
Electric field due to q₁ at distance (3-d) E1 = 14pivarepsilon0q1(3-d)2
Electric field due to q₂ at distance d E2 = 14pivarepsilon0q2d2
At point P where E = 0 E1 = E2
Working
Setting equal frac{16.0 × 10-6{(3-d)2} = frac{4.0 × 10-6{d2}
Simplifying 16.04.0 = (3-d)2d2
Taking square root sqrt{4} = 3-dd
Solving 2 = 3-dd
Algebraic manipulation 2d = 3 - d
Further simplification 3d = 3
Result
Distance from q₁ d = 1 m
12.3.Two opposite point charges, each of magnitude q are separated by a distance 2d. What is the electric potential at a point P mid-way between them?
Given
Distance between charges 2d
Positive charge +q
Negative charge -q
Formula
Potential due to positive charge V^+ = 14pivarepsilon0qd
Potential due to negative charge V^- = 14pivarepsilon0-qd
Total potential V = V^+ + V^- = 14pivarepsilon0qd + 14pivarepsilon0-qd
Result
Net potential V = 0
12.4.A particle carrying a charge of 2e falls through a potential difference of 3.0 V. Calculate the energy acquired by it.
Given
Magnitude of charge q = 2e
Potential difference V = 3.0 V
Formula
Energy acquired ΔK.E = qΔV
Working
Substituting values ΔK.E = (2e)(3.0 V)
In electron volts ΔK.E = 6 eV
Converting to Joules ΔK.E = 6.0 × 1.6 × 10-19 J
Result
Energy acquired ΔK.E = 9.6 × 10-19 J
12.5.In Millikan oil drop experiment, an oil drop of mass 4.9 × 10⁻¹⁵ kg is balanced and held stationary by the electric field between two parallel plates. If the potential difference between the plates is 750 V and the spacing between them is 5.0 mm, calculate the charge on the droplet. Assume g = 9.8 ms⁻².
Given
Mass of droplet m = 4.9 × 10-15 kg
Potential difference V = 750 V
Distance between plates d = 5.0 mm = 5.0 × 10-3 m
Value of g g = 9.8 m/s2
Formula
Equilibrium condition q = mgdV
Working
Substituting values q = frac{4.9 × 10-15 × 9.8 × 5.0 × 10-3{750}
Result
Charge on droplet q = 3.2 × 10-19 C
12.6.The time constant of a series RC circuit is t = RC. Verify that an ohm times farad is equivalent to second.
Formula
Ohm's law V = IR
Current definition I = qt
Working
Substituting V = qtR
Rearranging R = V × tq
Formula
Capacitance definition q = CV, quad C = qV
Multiplying R and C RC = V × tq × qV = t
Result
Unit verification 1 ohm × 1 farad = 1 second