Coulomb's law states that the force between two point charges varies as
Answer: A
The force is \( F = Q_1 Q_2/(4\pi\epsilon_0 R^2) \), directed along the line joining the charges and repulsive for like signs. The inverse square dependence is what makes Gauss's law take so simple a form in three dimensions.
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The value of \( 1/(4\pi\epsilon_0) \) in SI units is about
Answer: B
Correct answer: \( 9 \times 10^{9} \)
With \( \epsilon_0 = 8.85\times10^{-12} \) F/m, \[ \frac{1}{4\pi\epsilon_0} = 8.99\times10^{9} \approx 9\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}. \] The 8.85 offered is the permittivity itself rather than the constant built from it, \( 3\times10^{8} \) is the speed of light, and \( 4\pi\times10^{-7} \) is the permeability of free space.
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The permittivity of free space has units of
Answer: C
It appears in the capacitance of a parallel plate capacitor as \( \epsilon_0 A/d \), where the area over the separation is a length, so the constant itself must be a capacitance per unit length. Its value is \( 8.854 \times 10^{-12} \) F/m.
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The force between two charges of the same sign is
Answer: D
Coulomb's law gives a positive product for like charges and the vector form then points each charge away from the other. The force acts along the line of centres, which is what makes it a central force and conserves angular momentum about either charge.
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When several charges act on a test charge, the resultant force is obtained by
Answer: A
Superposition holds because Maxwell's equations are linear, so each source contributes independently. The vector nature matters: two equal and opposite contributions cancel entirely, which is how a field-free point arises between two like charges.
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Doubling the separation of two point charges changes the force between them to
Answer: B
Correct answer: a quarter of its former value
Coulomb’s law goes as the inverse square, \( F \propto 1/r^2 \), so doubling the separation divides the force by \( 2^2 = 4 \). The eighth offered would apply to an inverse cube, which is the dipole field rather than the point charge force - a distinction the dipole item in this topic makes explicitly.
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Electric field intensity at a point is defined as
Answer: C
Writing \( \mathbf{E} = \mathbf{F}/Q \) removes the test charge from the description and leaves a property of the source alone. The work per unit charge from infinity is the potential, which is a different and scalar quantity.
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The unit of electric field intensity is
Answer: D
Newtons per coulomb and volts per metre are the same unit, which follows from the joule being a newton metre and the volt a joule per coulomb. The second form is the one used in practice because potential differences are what get measured.
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The electric field at a distance \( R \) from a point charge \( Q \) has magnitude
Answer: A
Dividing Coulomb's force by the test charge leaves this expression, directed radially outward for a positive charge. The three wrong answers are each a correct formula for something else, which is why keeping the geometry in mind matters more than memorising forms.
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The field of an infinite straight line charge of density \( \rho_L \) at a distance \( \rho \) is
Answer: B
Applying Gauss's law to a cylinder round the line gives an area proportional to the radius, so the field falls as the first power rather than the square. It points radially away from the line and has no component along it.
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