Category: HC Verma Part 2: Electricity
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HC Verma Chapter 29 Problem 40 – Potential at Points Due to Dipole
Problem Statement An electric dipole has charges $\pm2.0\times10^{-8}$ C separated by 4.0 cm. Find the potential at a point (a) 3 cm from the positive charge along the axis, and (b) at the centre of the dipole. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics…
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HC Verma Chapter 29 Problem 39 – Electric Potential Energy of System of Three Charges
Problem Statement Three charges $+1\mu$C, $+2\mu$C, and $-3\mu$C are placed at vertices of an equilateral triangle of side 2 m. Find the total electric potential energy of the system. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles to the scenario described. The solution requires…
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HC Verma Chapter 29 Problem 38 – Potential at Centre of Square Charge Configuration
Problem Statement Four charges $q_1 = +1\mu$C, $q_2 = +2\mu$C, $q_3 = -3\mu$C, $q_4 = +4\mu$C are placed at the four corners of a square of side 1 m. Find the electric potential at the centre of the square. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies…
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HC Verma Chapter 29 Problem 37 – Superposition of Coulomb Forces
Problem Statement Solve the oscillation/wave problem: Four charges $+q$, $+q$, $-q$, $-q$ are placed at the four corners of a square of side $a$. The positive charges are at diagonally opposite corners. Find the force on any one of the charges. Vector superposition of Coulomb forces Step 1: Consider $+q$ at corner A. The other…
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HC Verma Chapter 29 Problem 36 – Electron in Electric Field
Problem Statement Determine the electric field for the configuration described: An electron is accelerated through a potential difference of 200 V. Find its final speed. (Electron: $m = 9.11\times10^{-31}$ kg, $q = 1.6\times10^{-19}$ C) Work-energy theorem: $qV = \frac{1}{2}mv^2$ Step 1: Energy gained: $K = qV = 1.6\times10^{-19}\times200 = 3.2\times10^{-17}$ J. Step 2: $$\frac Given…
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HC Verma Chapter 29 Problem 35 – Finding Charge from Potential Difference
Problem Statement Two concentric conducting spheres have radii 10 cm and 20 cm. The inner sphere has charge $Q$ and the outer is grounded. Find $Q$ if the potential of the inner sphere is 1000 V. Given Information See problem statement for all given quantities. Physical Concepts & Formulas This problem applies fundamental physics principles…
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HC Verma Chapter 29 Problem 34 – Electric Field Inside and Outside a Charged Shell
Problem Statement Determine the electric field for the configuration described: A thin conducting spherical shell of radius $R$ carries charge $Q$. Find the electric field (a) outside at distance $r > R$, (b) inside at $r Gauss’s law for spherical symmetry (a) Outside ($r > R$): Gaussian sphere of radius $r$ encloses charge $Q$: $$E\cdot4\pi…
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HC Verma Chapter 29 Problem 33 – Energy in Electric Field
Problem Statement Determine the electric field for the configuration described: Find the energy stored in the electric field between the plates of a capacitor with $C = 5\mu$F charged to $V = 100$ V. Energy stored: $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$ Step 1: $C = 5\times10^{-6}$ F, $V = 100$ V. Step 2:…
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HC Verma Chapter 29 Problem 32 – Charge on Parallel Plate Capacitor
Problem Statement Solve the capacitor/capacitance problem: A parallel plate capacitor with plate area $A = 0.02$ m$^2$ and separation $d = 2$ mm is connected to a 12 V battery. Find the charge on each plate. ($\varepsilon_0 = 8.85\times10^{-12}$ F/m) Capacitance: $C = \varepsilon_0 A/d$ Charge: $Q = CV$ Step 1: $$C = \frac{\varepsilon_0 A}{d}…
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HC Verma Chapter 29 Problem 31 – Capacitance of Isolated Sphere
Problem Statement Solve the capacitor/capacitance problem: Find the capacitance of an isolated conducting sphere of radius $R$. Capacitance: $C = Q/V$ Potential of sphere: $V = kQ/R$ Step 1: When charge $Q$ is given to the sphere, potential at surface: $$V = \frac{kQ}{R} = \frac{Q}{4\pi\varepsilon_0 R}$$ Step 2: $$C = \frac{Q}{V} = \frac{Q}{Q/(4\pi\varepsil Given Information…