When two conducting spheres are joined by a wire, charge flows until both reach the SAME potential (V = kQ/R). This makes charge share as Q proportional to R, so surface charge density becomes sigma proportional to 1/R. The smaller sphere ends up with the HIGHER surface charge density. Memory hook: "small sphere, sharp charge" — sigma1/sigma2 = R2/R1.
Two spheres joined by a wire reach the same potential, so charge shares as Q proportional to R while surface charge density sigma becomes proportional to 1/R — the smaller sphere carries the higher sigma.
Your doubts, answered
When two spheres are connected by a wire, do they get equal charge or equal potential?
Equal POTENTIAL, not equal charge. The wire lets charge move freely until the potential is the same on both, because a difference in potential would keep pushing charge through the wire. So we set V1 = V2, that is kQ1/R1 = kQ2/R2. This gives Q1/Q2 = R1/R2. The bigger sphere holds MORE charge, but they share the same V.
Why is the surface charge density higher on the SMALLER sphere?
Charge shares as Q proportional to R (bigger sphere, more charge). But surface area grows as R squared. So sigma = Q/(4 pi R squared) behaves like R/R squared = 1/R. Since sigma is proportional to 1/R, the smaller radius gives the LARGER sigma. The charge crowds more tightly on the small sphere.
Is it Q that is inversely proportional to R, or sigma?
Charge Q is DIRECTLY proportional to R (Q proportional to R). Surface charge density sigma is INVERSELY proportional to R (sigma proportional to 1/R). Do not mix them. NEET often traps students who write Q proportional to 1/R. Remember: equal potential fixes the charge ratio as R, then dividing by area R squared flips it to 1/R for density.
What is the connection to lightning rods and sharp points?
Same physics. A sharp point is like a very small radius sphere, so it has very high sigma and a very strong field just outside (E = sigma/epsilon0). This is why charge leaks and sparks form at sharp tips. This 'action of points' is the real-world version of the connected-spheres result and is directly asked in NEET reasoning questions.
⚠️ The NEET trap ✗ Assuming the two connected spheres end up with EQUAL charge (Q1 = Q2), so equal sigma too. ✓ They reach equal POTENTIAL, so Q proportional to R and sigma proportional to 1/R. The smaller sphere has higher sigma; sigma1/sigma2 = R2/R1. 🧠 Wire equalises VOLTAGE, not charge. Small sphere = sharp charge (higher sigma).
Real NEET questions
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: A wire connects them, so they reach a COMMON potential. V = kQ1/R1 = kQ2/R2, which gives Q1/Q2 = R1/R2. Step 2: Surface charge density sigma = Q/(4 pi R^2). So sigma1/sigma2 = (Q1/R1^2)/(Q2/R2^2) = (Q1/Q2)(R2^2/R1^2). Step 3: Substitute Q1/Q2 = R1/R2: sigma1/sigma2 = (R1/R2)(R2^2/R1^2) = R2/R1. So sigma is proportional to 1/R. Answer: (D) R2/R1.
2019
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 will be the new surface charge densities on them?
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). Total Q = 5 sigma(4 pi R^2). Step 2: On contact they reach a common potential, so charge splits in the ratio of radii, Q1':Q2' = R:2R = 1:2. Q1' = (1/3)(5 sigma 4 pi R^2) = (5sigma/3)(4 pi R^2). Q2' = (2/3)(5 sigma 4 pi R^2) = (10sigma/3)(4 pi R^2). Step 3: New densities. sigma1 = Q1'/(4 pi R^2) = 5sigma/3. sigma2 = Q2'/(4 pi (2R)^2) = (10sigma/3)/4 = 5sigma/6. Answer: (D) sigma1 = 5sigma/3, sigma2 = 5sigma/6 (smaller sphere has higher density).
Solved Electric Charges And Fields NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for surface charge density of connected spheres?
For two spheres joined by a wire, sigma1/sigma2 = R2/R1, that is sigma is inversely proportional to R. This comes from equal potential (Q proportional to R) divided by area (R squared).
Does the bigger sphere have more charge or more surface charge density?
The bigger sphere has MORE total charge (Q proportional to R) but LESS surface charge density (sigma proportional to 1/R). The smaller sphere has the higher sigma.
Why do connected conductors reach the same potential?
The wire lets charge move freely. Any potential difference pushes charge through the wire until the potentials are equal, so we always set V1 = V2 for connected conductors.
Is this the same as charge sharing between two conductors?
Yes, it is the same physics. Charge sharing gives the charge ratio (Q proportional to R); this concept just takes the extra step of finding the surface charge density sigma = Q/area.
How is this linked to lightning rods?
A sharp tip acts like a tiny-radius sphere, so it has very high sigma and a strong field (E = sigma/epsilon0). This makes charge escape at points, which is why lightning rods are pointed.
