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Q. No.An electron falls through a distance of \(1.5\) cm in a uniform electric field of magnitude \(2\times10^4\) N/C [figure (a)]. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [figure (b)]. If \(t_e\) and \(t_p\) are the time of fall for electron and proton respectively, then:
1. \(t_e=t_p\)
2. \(t_e>t_p\)
3. \(t_e<t_p\)
4. none of these
Subtopic: Â Electric Field |
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Three charges \(q,~q,~-q\) are placed at the three corners of an equilateral triangle \(ABC\), of side \(a.\)
The mid-point of side \(AB\) is \(P\) while the circumcenter of \(ABC\) is \(O\). Let the electric field at \(P\) be \(E_p\) and that at \(O\) be \(E_O.\)
Then, \(E_O:E_P=\)
1. \(\dfrac{2}{9}\)
2. \(\dfrac{4}{9}\)
3. \(\dfrac{9}{2}\)
4. \(\dfrac{9}{4}\)
The mid-point of side \(AB\) is \(P\) while the circumcenter of \(ABC\) is \(O\). Let the electric field at \(P\) be \(E_p\) and that at \(O\) be \(E_O.\)
Then, \(E_O:E_P=\)
1. \(\dfrac{2}{9}\)
2. \(\dfrac{4}{9}\)
3. \(\dfrac{9}{2}\)
4. \(\dfrac{9}{4}\)
Subtopic: Â Electric Field |
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The figure shows some of the electric field lines corresponding to an electric field. The figure suggests that:
1. \(E_A>E_B>E_C\)
2. \(E_A=E_B=E_C\)
3. \(E_A=E_C>E_B\)
4. \(E_A=E_C<E_B\)
Subtopic: Â Electric Field |
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Which of the following field configurations is/are possible?
Note: \(A,B,C\) are conductors. Other charges may be present in the vicinity.
Note: \(A,B,C\) are conductors. Other charges may be present in the vicinity.
| 1. | I, III | 2. | II |
| 3. | I, II, III | 4. | none of I, II, III |
Subtopic: Â Electric Field |
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Identical point charges (\(q\) each), are placed at the eight corners of a cube of side \(a.\) When one of the charges is removed, the electric field at the center becomes \(E_c.\)
Now, identical point charges (same magnitude \(q\) each), are placed at the four corners of a square of side \(a.\) When one of the charges is removed, the electric field at the center becomes \(E_s.\) Then,
1. \(\dfrac{E_s}{2}=\dfrac{E_C}{3}\)
2. \(\dfrac{E_s}{3}=\dfrac{E_C}{2}\)
3. \(\dfrac{E_s}{\sqrt2}=\dfrac{E_C}{\sqrt3}\)
4. \(\dfrac{E_s}{\sqrt3}=\dfrac{E_C}{\sqrt2}\)
Now, identical point charges (same magnitude \(q\) each), are placed at the four corners of a square of side \(a.\) When one of the charges is removed, the electric field at the center becomes \(E_s.\) Then,
1. \(\dfrac{E_s}{2}=\dfrac{E_C}{3}\)
2. \(\dfrac{E_s}{3}=\dfrac{E_C}{2}\)
3. \(\dfrac{E_s}{\sqrt2}=\dfrac{E_C}{\sqrt3}\)
4. \(\dfrac{E_s}{\sqrt3}=\dfrac{E_C}{\sqrt2}\)
Subtopic: Â Electric Field |
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The electric field, at the centre of a square with charges placed at its four vertices as shown in the figure, is: \(\Big(k=\dfrac{1}{4\pi\varepsilon_0}\Big)~\)
1. zero
2. \(4\dfrac{kq}{a^2}\)
3. \(2\dfrac{kq}{a^2}\)
4. \(2\sqrt2\dfrac{kq}{a^2}\)
1. zero
2. \(4\dfrac{kq}{a^2}\)
3. \(2\dfrac{kq}{a^2}\)
4. \(2\sqrt2\dfrac{kq}{a^2}\)
Subtopic: Â Electric Field |
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Four charges \(q,q\) and \(-q,-q\) are placed at the four vertices of a square of side \(a,\) with like charges across a diagonal. The electric field at the centre of the square is: \((k=\frac{1}{4\pi\varepsilon_0})\)
1. zero
2. \(\sqrt2\frac{kq}{a^2}\)
3. \(2\frac{kq}{a^2}\)
4. \(4\frac{kq}{a^2}\)
1. zero
2. \(\sqrt2\frac{kq}{a^2}\)
3. \(2\frac{kq}{a^2}\)
4. \(4\frac{kq}{a^2}\)
Subtopic: Â Electric Field |
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