#potential difference

Conductors, Electrostatic Shielding and Dielectrics

In a system where a charged particle moves along a curved path between two points with different potentials, what can be

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. The work done by the electric field on a charge q moving between two points is W = q (Vfiₙₐl - Viₙitiₐl) , where V is the potential. Since the electrostatic field is conservative, this work depends only on the potential difference between the points and not on the path taken (straight or curved). Thus, the work done is path-independent and determined solely by the

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Sharing of Charges, Common Potential and Applications

In a parallel combination of capacitors, why does each capacitor have the same potential difference across its plates?

**Sharing of charges** when charged capacitor C₁ at V₁ connected to uncharged C₂, total charge Q = C₁ V₁ conserved, common potential V_common = Q/(C₁+C₂) = C₁ V₁/(C₁+C₂), final charges Q₁' = C₁ V_common, Q₂' = C₂ V_common. For 4 μF at 100 V (Q=4×10⁻⁴ C) connected to 4 μF uncharged, V_common=4×10⁻⁴/8×10⁻⁶=50 V. In a parallel combination, the capacitors are connected across the same two nodes of the circuit. Since voltage (potential difference) is the same between these nodes (as they are directly connected to the same battery or voltage source), each capacitor experiences the same potential difference across its plates. The charge on each

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Combination of Capacitors - Series and Parallel

In a series combination of capacitors, why does each capacitor have the same charge but different potential differences?

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. In a series combination, capacitors are connected end-to-end, forming a single path for charge flow. When a voltage is applied, the same charge Q accumulates on each capacitor (as charge conservation ensures the same Q passes through each during charging). However, the potential difference across each capacitor is V = Q/C , so V

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Wheatstone Bridge and Meter Bridge

In a Wheatstone bridge, when the bridge is balanced, what is the potential difference across the galvanometer?

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. At balance, the potential at the two junctions connected to the galvanometer is equal (due to the ratio condition R₁ / R₂ = R₃ / R₄ ), so the potential difference across the galvanometer is zero, and no current flows through it. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P =

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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