Physics · Electrostatic Potential And Capacitance · NEET
When two charged conducting spheres of radii R1 and R2 are joined by a wire, they reach the same potential, and the surface charge density is inversely proportional to the radius: sigma is proportional to 1/R, so sigma1/sigma2 = R2/R1. The smaller sphere ends up with the larger charge density. Memory hook: "Small sphere, sharp charge" - a smaller radius packs charge more tightly, which is exactly why lightning rods are made pointed.
Two conductors joined by a wire settle at one common potential. Charge follows radius (Q proportional to R), but density follows 1/R, so the smaller sphere has the denser, sharper charge (shown by longer field arrows).
Your doubts, answered
When two spheres are connected by a wire, do they have the same charge or the same potential?
They share the SAME potential, not the same charge. The wire lets charge flow until both spheres sit at one common potential V = kQ/R. Because V must be equal but R is different, the charges Q1 and Q2 come out different: Q1/Q2 = R1/R2. So the bigger sphere carries more total charge, but as you will see, the smaller sphere has the bigger surface charge density.
Why does the smaller sphere have a larger surface charge density?
Start from equal potential: kQ1/R1 = kQ2/R2, so Q is proportional to R. Now sigma = Q/(4 pi R^2), which is proportional to R/R^2 = 1/R. So sigma is inversely proportional to R. A smaller radius means a larger sigma. In plain words, the same pull of potential forces charge into a smaller surface, so it becomes more crowded there.
How do I remember sigma1/sigma2 = R2/R1 and not R1/R2?
The ratio flips because sigma is inversely proportional to R. Sphere 1 has radius R1, and its density sigma1 is proportional to 1/R1. So sigma1/sigma2 = (1/R1)/(1/R2) = R2/R1. The subscripts swap. Quick check: if R1 is smaller, then sigma1 should be bigger, and R2/R1 is indeed bigger than 1 - it matches.
How is this connected to why the field is stronger at sharp points?
Just outside a conductor, E = sigma/epsilon0, so E is proportional to sigma. Since sigma is proportional to 1/R, the field is strongest where the radius of curvature is smallest - the sharp points and edges. This is the corona discharge and lightning rod effect. NCERT states this directly: charge density is higher at sharper, more pointed regions of a conductor.
⚠️ The NEET trap ✗ sigma1/sigma2 = R1/R2 (thinking density follows radius directly, since the bigger sphere has more charge). ✓ sigma1/sigma2 = R2/R1, because sigma is proportional to 1/R. The bigger sphere has more total charge but a smaller charge density. 🧠 NTA loves swapping this ratio. Total charge follows R (Q proportional to R), but surface density follows 1/R. Always divide the charge by the area 4 pi R^2 before comparing - the extra R^2 flips the ratio.
Real NEET questions
NEET 2021
Two charged spherical conductors of radii R1 and R2 are connected by a wire. Then the ratio of the surface charge densities of the spheres (sigma1/sigma2) is
A · (R1/R2)^2
B · R2^2/R1^2
C · R1/R2
D · R2/R1 ✓
Solution: Step 1: The wire connects the spheres, so they reach a COMMON potential. V = kQ1/R1 = kQ2/R2. Step 2: This gives Q1/Q2 = R1/R2 (charge follows radius). Step 3: Surface charge density sigma = Q/(4 pi R^2). So sigma1/sigma2 = (Q1/R1^2)/(Q2/R2^2) = (Q1/Q2) x (R2^2/R1^2). Step 4: Substitute Q1/Q2 = R1/R2: sigma1/sigma2 = (R1/R2) x (R2^2/R1^2) = R2/R1. Therefore sigma is proportional to 1/R, and the answer is (d) R2/R1.
NEET 2019 (Odisha)
Two metal spheres, one of radius R and the other of radius 2R, have the same surface charge density sigma. They are brought in contact and then separated. What are the new surface charge densities?
A · sigma1 = 5sigma/6, sigma2 = 5sigma/2
B · sigma1 = 5sigma/2, sigma2 = 5sigma/6
C · sigma1 = 5sigma/2, sigma2 = 5sigma/3
D · sigma1 = 5sigma/3, sigma2 = 5sigma/6 ✓
Solution: Step 1: Initial charges Q = sigma x area. Q1 = sigma(4 pi R^2). Q2 = sigma(4 pi (2R)^2) = 4 sigma(4 pi R^2). Step 2: Total charge Q = Q1 + Q2 = 5 sigma(4 pi R^2), conserved on contact. Step 3: On contact both reach common potential, so charge splits in proportion to radius: Q1':Q2' = R:2R = 1:2. So Q1' = (1/3)(5 sigma 4 pi R^2) and Q2' = (2/3)(5 sigma 4 pi R^2). Step 4: New densities. sigma1 = Q1'/(4 pi R^2) = 5 sigma/3. sigma2 = Q2'/(4 pi (2R)^2) = (10 sigma/3)/4 = 5 sigma/6. Answer (d): sigma1 = 5sigma/3, sigma2 = 5sigma/6. Note the smaller sphere again ends with the higher density.
Solved Electrostatic Potential And Capacitance NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Is surface charge density directly or inversely proportional to radius for connected spheres?
Inversely proportional. sigma is proportional to 1/R. So the smaller sphere has the larger surface charge density, even though the larger sphere holds more total charge.
What is the surface charge density ratio for two connected spheres?
sigma1/sigma2 = R2/R1. The subscripts are swapped compared with the radii because density goes as 1/R.
What stays equal when two conductors are joined by a wire - charge or potential?
Potential. Charge flows until both conductors sit at one common potential V = kQ/R. The individual charges then differ so that this common potential is satisfied.
How does this concept explain a lightning rod?
The field just outside a conductor is E = sigma/epsilon0, and sigma is proportional to 1/R. At a sharp point the radius of curvature is tiny, so sigma and E are huge. This strong field ionises the air (corona discharge), which is why lightning rods are pointed.
Does the total charge stay conserved when spheres are connected?
Yes. Connecting by a wire only redistributes charge; the total Q1 + Q2 before equals the total after. Potential becomes equal, but total charge is conserved.
