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● Physics · Class 11 · 2.3 Potential due to a point charge
What does V = q / (4πε₀r) actually tell me?
V
It says the potential falls off as 1/r: double the distance from the charge and the potential halves. Think of it as the “energy per unit charge” needed to bring a test charge in from infinity.
V
◎ CHECKPOINT
If q is negative, what is the sign of V at any point near it — and why?
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📖 2.3 Potential due to a point charge · Unit 1 of 2Next lesson 🔒
Potential at a distance r

Bring a unit positive test charge in from infinity. The work you do against the field is the potential at that point — and for a point charge it depends only on how far away you stop:

V(r) = q / 4πε₀r
◎ ANSWER THE QUESTION TO CONTINUE
Explain why V is negative near a negative charge.
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Question 1 of 5
A 2 × 10⁻⁸ C point charge sits at the origin. What is the potential 3 cm away?
A1.8 × 10⁵ V
B6 × 10³ V
C6 × 10⁴ V
D1.8 × 10⁴ V
Correct! V = kq/r = (9 × 10⁹)(2 × 10⁻⁸)/0.03 = 6 × 10³ V.

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Just a hintWalk me through itFull solution
📷  2 kg block · μ = 0.2 · 20 N at 30° — find the acceleration
Step 1 — Resolve the force
Fₓ = F cos 30° = 17.3 N and F_y = F sin 30° = 10 N
Step 2 — Find the normal force
N = mg − F_y = 20 − 10 = 10 N
Step 3 — Friction, then a
f = μN = 2 N, so a = (17.3 − 2)/2 ≈ 7.66 m/s²

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