IBDP Physics- D.2 Electric and magnetic fields- IB Style Questions For SL Paper 1A -FA 2025
▶️ Answer/Explanation
1. Speed of Light Relationship:
\(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\).
Therefore, \(\mu_0 \epsilon_0 = \frac{1}{c^2}\).
2. Units of Speed:
Units of \(c\) = m s\(^{-1}\).
3. Units of \(\mu_0 \epsilon_0\):
Units = \(\frac{1}{(m s^{-1})^2} = \frac{1}{m^2 s^{-2}} = m^{-2} s^2\).
✅ Answer: (A)
▶️ Answer/Explanation
For two positive charges, the electric field vectors point away from each charge.
At positions outside the pair (such as \(A\) and \(D\)), both fields point in the same direction, so they cannot cancel to give zero.
A zero field point (if it exists) must be between the two charges where the fields are in opposite directions.
Between the charges, set magnitudes equal:
\(\displaystyle E_q = E_{2q}\Rightarrow \frac{kq}{x^2}=\frac{k(2q)}{(d-x)^2}\).
This requires \((d-x)^2 = 2x^2\), so \((d-x) > x\) and the zero-field point is closer to the smaller charge \(q\).
From the labelled positions, this corresponds to \(B\).
✅ Answer: (B)
▶️ Answer/Explanation
Consider the electric field at the empty corner due to each charge (field points away from \(+Q\) and towards \(-Q\)).
Let the square side be \(a\). Then two of the charges are at distance \(a\) from the field point, and the diagonal charge is at distance \(\sqrt{2}\,a\).
The fields from charges at distance \(a\) have magnitude \(\displaystyle E_a=\frac{kQ}{a^2}\) (or \(\displaystyle \frac{k|Q|}{a^2}\)), directed along the sides toward/away from the charges.
The field from the diagonal charge has smaller magnitude \(\displaystyle E_d=\frac{kQ}{(\sqrt{2}a)^2}=\frac{kQ}{2a^2}\), directed along the diagonal.
Adding the side-components and the diagonal-component (vector addition) gives a resultant direction corresponding to option B.
✅ Answer: (B)