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Q.Two point charges q₁ = 3 μC and q₂ = −3μC are located 20cm apart in vacuum.

i) What is the electric field at the mid-point of the line joining the two charges?
ii) If a test charge of magnitude 1.5 × 10⁻⁹C is placed at this point, what is the force experienced by this charge? OR Find the equivalent capacitance of the capacitors in the network shown below. Also, calculate the total charge in the network when a 100V battery is connected across the combination. Given that C₁=C₅= 1 μF, C₂=C₃=C₄ =2 μF.
Nagaland NbseNagaland Board of School Education 2025Subjective· 3mImportance★★★★★
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Both charges' fields point the same way at the midpoint (away from +q₁, toward −q₂), so they add directly to 5.4×10⁶ N/C; the force on the test charge follows from F=qE.

Given: q1=3 μC=3×10−6 Cq_1 = 3\ \mu\text{C} = 3\times10^{-6}\ \text{C}, q2=−3 μCq_2 = -3\ \mu\text{C}, separation =20 cm=0.2 m= 20\ \text{cm} = 0.2\ \text{m}, so distance from midpoint to each charge, r=0.1 mr = 0.1\ \text{m}.

i) Field at the midpoint:

Field due to q1q_1 (positive) at the midpoint points away from q1q_1, i.e. towards q2q_2:

E1=kq1r2=9×109×3×10−6(0.1)2=2.7×1040.01=2.7×106 N/CE_1 = \frac{kq_1}{r^2} = \frac{9\times10^9 \times 3\times10^{-6}}{(0.1)^2} = \frac{2.7\times10^4}{0.01} = 2.7\times10^6\ \text{N/C}

Field due to q2q_2 (negative) at the midpoint points towards q2q_2 (field lines converge into a negative charge) — the same direction as E1E_1:

E2=k∣q2∣r2=2.7×106 N/CE_2 = \frac{k|q_2|}{r^2} = 2.7\times10^6\ \text{N/C} (same direction as E1E_1)

Since both fields point the same way (from q1q_1 towards q2q_2), they add: …

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