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The unit of charge is Coulombs. It is a scalar quantity because it represents magnitude only and does not have a direction.

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Milikan's oil drop experiment showed the existence of the fundamental charges, specifically the charge of an electron.

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An electric field is a region where an electric force is experienced by a charged particle.

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The electric field is \( \frac{F}{q} \), so A and B are wrong. D is the force experienced by a charge in a field, and strength of the field is the force per unit charge, so \( F = k \frac{Q}{r^2} \) would be a correct expression, however, the density of field lines in a given area also shows the strength of a field.

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Electric field has 2 equations, one with the force, \( E = \frac{F}{q} \) which has units \( \frac{N}{C} \), and one from the field between 2 parallel plates which is \( E = \frac{V}{d} \), which has units \( \frac{V}{m} \), hence B.

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That is important to remember. Magnetic fields are not like electric fields, they do not have a starting point and an ending point. Outside the magnet the field lines move from north to south, but to complete the loop the field lines inside the magnet itself must flow from south to north. Since the field lines are the most dense inside the magnet, that is where the field is the strongest.

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The work done is:

\[ W = qEd \] Substituting: \[ W = (1.5 \times 10^{-6})(300)(0.02) = 9.0 \times 10^{-6} \, \text{J}. \] Since the charge moves opposite to the field direction, the work done is negative: \[ W = -9.0 \cdot 10^{-6} \, \text{J}. \]

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For the droplet to remain motionless, the electric force \( F_e \) must equal the gravitational force \( F_g \).

\[ F_e = qE \quad \text{and} \quad F_g = mg \]

The electric field \( E \) between the plates is given by:

\[ E = \frac{V}{d} = \frac{300}{0.005} = 6.0 \times 10^4 \, \text{N}/\text{C} \]

Thus, \( F_e = F_g = qE \):

\[ F_g = (1.6 \times 10^{-19})(6.0 \times 10^4) = 9.6 \times 10^{-15} \, \text{N} \]

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The force on a charge in an electric field is given by: \[ F = qE \]

Thus, the correct answer is C.

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For a hollow spherical conductor:

• The electric potential is constant throughout the hollow region because there is no electric field inside the conductor (from Gauss's law).
• The electric field is zero everywhere inside the hollow region, as the net charge on the conductor resides on its outer surface, and the enclosed charge inside the hollow is zero.

Thus, the correct answer is A.

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The force between two charges is given by Coulomb's law: \[ F = k \frac{q_1 q_2}{r^2} \] Since the charges remain the same, the force is inversely proportional to the square of the distance: \[ F' = F \left(\frac{r}{r'}\right)^2 \] Substituting \( F = 18 \:\mathrm{N}, r = 2 \:\mathrm{m}, r' = 6 \:\mathrm{m} \): \[ F' = 18 \left(\frac{2}{6}\right)^2 = 18 \times \frac{1}{9} = 2 \:\mathrm{N} \]

Thus, the correct answer is B.

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The work done in moving a charge \( q \) between two points is given by: \[ W = q \Delta V \] Where \( \Delta V = V_B - V_A \) is the change in electric potential.

1. From \( A \) to \( B \): \[ \Delta V = V_B - V_A = -25 - (+15) = -40 \:\mathrm{V} \] \[ W_{A \to B} = (-4 \:\mathrm{C})(-40 \:\mathrm{V}) = 160 \:\mathrm{J} \] 2. Total work done (moving \( A \to B \) and back to \( A \)): Since the work done \( A \to B \) is equal and opposite to the work done \( B \to A \), the total work is: \[ W_{\text{total}} = W_{A \to B} + W_{B \to A} = 160 \:\mathrm{J} - 160 \:\mathrm{J} = 0 \:\mathrm{J} \]

Thus, the correct answer is A.

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The electric potential due to a single point charge is given by: \[ V = \frac{kQ}{r} \] where \( k \) is the Coulomb constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the potential is calculated. At the center of the square (1) the distance \( r \) from each charge to the center is the same and can be calculated using the geometry of the square: \[ r = \frac{\sqrt{2}}{2} \cdot (3d) = \frac{3d}{\sqrt{2}} \] (2) The total potential at the center is the sum of the potentials due to the four charges: \[ V_{\text{total}} = 4 \cdot \frac{kQ}{r} = 4 \cdot \frac{kQ}{\frac{3d}{\sqrt{2}}} = \frac{4kQ \sqrt{2}}{3d} \]

Thus, the correct answer is B.

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Correct answer: B

Electric field lines are an example of a model to aid in visualizing electric fields. They help us understand the direction and strength of the electric field in space around charges by showing the direction of the force that a positive test charge would experience. They do not represent actual physical lines but are a conceptual tool used in electrostatics.