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The electric force between two charged particles is always directed from the positive particle to the negative particle. This is because opposite charges attract each other, while like charges repel each other.

When a negatively charged rod is brought close to a metal sphere, the free electrons in the sphere are repelled from the rod and move to the other end of the sphere. This creates a region of positive charge on the side of the sphere closest to the rod, and a region of negative charge on the opposite side. The process of charge distribution stops when the net force on the free electrons inside the metal is equal to zero. If the sphere is then earthed, the free electrons will flow from the sphere to the ground, leaving the sphere with a net positive charge.

The angle of dip is the angle between the direction of the intensity of the total magnetic field of Earth and a horizontal line in the magnetic meridian. It is also known as the magnetic inclination. The angle of dip ranges from 0 degrees at the magnetic equator to 90 degrees at the magnetic poles.

The polarities at X and Y would be north and north.

Ep = 2.2kV = 2200V; Es=110V; Ns=25; Np=? =\(\frac{E_p}{E_s}\) = \(\frac{N_p}{N_s}\) = \(\frac{2200}{110}\) = \(\frac{N_p}{25}\) =110 \times Np = 2200 \times 25 =\(N_p\) = \(\frac{55,000}{110}\) ∴\(N_p\) = 500 turns

An AC generator uses slip rings to transfer the induced current smoothly to the circuit. A DC generator uses split rings to transfer the induced current to the circuit and also convert the induced AC into pulsating DC. So, to convert an AC generator into a DC generator, the slip rings needs to be replaced with split rings.

ε = 24.0 V, r=1\Omega, R=5.0\Omega terminal p·d = ? ε = IR+Ir = I(R+r) ⇒ IR+Ir = I(R+r) ⇒ I = \(\frac{24}{5+1}\) ⇒ I =\(\frac{24}{6}\) ⇒ I = 4 A terminal p·d = 4 \times 5 terminal p·d = 20.0 V

A= \(0.8m^2\) d= 20mm =\(\frac{20}{1000}\) = 0.02m v =120v; \(ε_oA\)= \(8.85 \times 10^-12 fm^-1\) C = \(\frac{ε_oA}{d}\) C =\(\frac{8.85 \times 10^-12\times0.8}{0.02}\) C= \(3.54 \times 10^-10 F\) Q= CV ⇒ \(3.54 \times 10^-10\) \times 120 = \(4.25\times10^-8c\) Q=\(42.5 \times 10^-9c\) = 42.5nC

q=\(4.6 \times 10^{-5}\), C= \(3.2 \times 10^4\) \(Vm^{-1}\) F= ? F=qE ⇒ \(4.6 \times 10^{-5}\)C \times \(3.2 \times 10^4\) ; F= 1.5 N

Electrical conduction of gas is applied in: (i) The identification of gases (ii) Lighting/fluorescent tubes (iii) Advertising industry/Neon signs (iv) Cathode ray oscilloscope/T.V. tubes

For the combination in series; ⇒R1 = 35k\Omega + 40k\Omega = 75k\Omega R is combined with 75k\Omega in parallel to give 25k\Omega = \(\frac{1}{R_eq}\) = \(\frac{1}{R}\) + \(\frac{1}{R}\) = \(\frac{1}{25}\) = \(\frac{1}{R}\) + \(\frac{1}{75}\) = \(\frac{1}{25}\) - \(\frac{1}{75}\) + \(\frac{1}{R}\) = \(\frac{3-1}{75}\) = \(\frac{1}{R}\) = \(\frac{2}{75}\) = \(\frac{1}{R}\) = \(\frac{75}{2}\) = R ; R = 37.5k \Omega

The electrolyte used in the Nickel-Iron (NiFe) accumulator is potassium hydroxide solution. It is a non-flammable, non-toxic, and relatively stable electrolyte. It is also a good conductor of electricity, which is important for the operation of the battery.

Rubber is a good insulator because it does not allow electrons to flow through it easily. Copper, silver, and water are all good conductors of electricity.

An incandescent light source is a type of light source that produces light by heating a material until it glows. The material that is heated is usually a metal, such as tungsten. Fluorescent lamps, LED lamps, and neon lamps are all examples of non-incandescent light sources. Fluorescent lamps produce light by passing an electric current through a gas, which causes the gas to emit ultraviolet light. The ultraviolet light then causes a coating on the inside of the lamp to fluoresce, which produces visible light. LED lamps produce light by passing an electric current through a semiconductor material, which causes the material to emit photons. The photons then produce visible light. Neon lamps produce light by passing an electric current through a gas, which causes the gas to emit visible light.

\(V_L\)=60 V ,\(V_C\) = 120 V , \(V_R\) =30V ;V = ? ; \(V^2\) = \(V_R^2\) + \(V_L\) - \(V_C^2\) = \(V^2\) = \(30^2\) + \(60 - 120^2\) = \(V^2\) = 900 + 3600 = \(V^2\) = 4500 = V = \sqrt4500 = 67V