Physics · Electrostatic Potential And Capacitance · NEET
When you fill the gap of a capacitor with a dielectric (an insulator like glass or mica), its capacitance goes UP by a factor K, the dielectric constant: C = K C0, where C0 is the air/vacuum value. The dielectric polarises and weakens the field inside, so the plates hold more charge at the same voltage. Memory hook: "dielectric = more charge stored = bigger C" (K is always greater than 1, so C only grows, never shrinks).
Left: vacuum gap has a strong field E0 and capacitance C0. Right: the same capacitor filled with a dielectric polarises, weakening the field to E0/K, so the same charge sits at a lower voltage and capacitance rises to C = K C0.
Your doubts, answered
Does adding a dielectric increase or decrease capacitance?
It always INCREASES it. The new capacitance is C = K C0, and the dielectric constant K is always greater than 1 (K = 1 only for vacuum). So a dielectric can only make capacitance bigger, never smaller. For water K is about 80, so C becomes 80 times larger.
Why does capacitance increase when a dielectric is inserted?
The dielectric polarises: its molecules line up and create a small opposite field inside. This reduces the net field E, so for the same charge the voltage V = E d drops. Since C = Q/V, a smaller V means a bigger C. The plates can now store more charge for the same voltage.
With the battery still connected, what changes when I insert a dielectric?
The battery keeps V FIXED. Capacitance rises to C = K C0, so charge rises: Q = C V = K C0 V. More charge flows in from the battery. Energy U = (1/2) C V^2 also increases K times. Remember: battery connected means V is constant, Q and U grow.
If the battery is removed first, then a dielectric is inserted, what changes?
Now the CHARGE Q is fixed (nowhere for it to go). Capacitance still rises to C = K C0, so voltage drops: V = Q/C = V0/K. The field E and energy U = Q^2/(2C) both drop by K. Remember: battery removed means Q is constant, V and U shrink.
Why does the electric field decrease inside a dielectric?
The external field pulls the dielectric's positive and negative charges slightly apart (polarisation). This creates bound surface charges that set up a field opposing the original one. The net field becomes E = E0/K, weaker than in vacuum. That weaker field is the whole reason capacitance goes up.
⚠️ The NEET trap ✗ Inserting a dielectric always increases the stored energy of the capacitor. ✓ It depends on whether the battery is connected. Battery connected (V fixed): energy increases K times, U = (1/2)K C0 V^2. Battery removed (Q fixed): energy DECREASES to U0/K, because U = Q^2/(2C) and C got bigger. 🧠 First ask: is the battery still connected? Battery on means V constant; battery off means Q constant. Decide that BEFORE touching any formula.
Real NEET questions
NEET 2020
The capacitance of a parallel plate capacitor with air as medium is 6 microF. With the introduction of a dielectric medium, the capacitance becomes 30 microF. The permittivity of the medium is (eps0 = 8.85 x 10^-12 C^2 N^-1 m^-2):
A · 0.44 x 10^-10 C^2 N^-1 m^-2 ✓
B · 5.00 C^2 N^-1 m^-2
C · 0.44 x 10^-13 C^2 N^-1 m^-2
D · 1.77 x 10^-12 C^2 N^-1 m^-2
Solution: Step 1: Find the dielectric constant. C = K C0, so K = C/C0 = 30/6 = 5. Step 2: Permittivity of the medium is eps = K eps0. Step 3: eps = 5 x 8.85 x 10^-12 = 44.25 x 10^-12 = 0.44 x 10^-10 C^2 N^-1 m^-2. Answer: A.
NEET 2016 (Phase 2)
A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials of constants k1, k2, k3 and k4 as shown (top half split into three equal areas over k1, k2, k3; bottom half is k4). For a single dielectric k giving the same C, k is given by:
A · k = k1 + k2 + k3 + 3k4
B · k = (2/3)(k1 + k2 + k3) + 2k4
C · 2/k = 3/(k1 + k2 + k3) + 1/k4 ✓
D · 1/k = 1/k1 + 1/k2 + 1/k3 + 3/(2k4)
Solution: Step 1: Top half (thickness d/2). k1, k2, k3 each cover area A/3 side by side, so they are in PARALLEL. Effective top constant k_top = (k1 + k2 + k3)/3. Step 2: The top half (d/2) and bottom half k4 (d/2) are stacked, so they are in SERIES. For two equal-thickness halves the equivalent K obeys 2/k = 1/k_top + 1/k4. Step 3: Substitute k_top: 2/k = 3/(k1 + k2 + k3) + 1/k4. Answer: C.
Solved Electrostatic Potential And Capacitance NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for the effect of a dielectric on capacitance?
C = K C0, where C0 is the capacitance with air or vacuum and K is the dielectric constant. For a parallel plate capacitor this is C = K eps0 A / d.
Can a dielectric ever reduce capacitance?
No. K is always greater than or equal to 1, so C = K C0 is always at least C0. A dielectric can only keep C the same (vacuum, K = 1) or increase it.
What is the maximum charge a dielectric-filled capacitor can hold?
It is limited by dielectric strength, the maximum field the dielectric can survive before breaking down and conducting. Beyond that field the dielectric fails, so a good dielectric also lets a capacitor work at higher voltage safely.
Why do real capacitors use a dielectric instead of air?
A dielectric gives more capacitance in the same size, lets the plates sit closer without touching, and raises the breakdown voltage. So it stores more charge in a smaller, safer package.
