- 1(current)
Q. No.Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is \(r\) (as shown in Fig. I). Now, as shown in Fig. II, the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:
1. \(\frac{r}{\sqrt[3]{2}}\)
2. \(\frac{r}{\sqrt[2]{2}}\)
3. \(\frac{2r}{3}\)
4. none of the above
1. \(\frac{r}{\sqrt[3]{2}}\)
2. \(\frac{r}{\sqrt[2]{2}}\)
3. \(\frac{2r}{3}\)
4. none of the above
Subtopic: Â Coulomb's Law |
 70%
From NCERT
AIPMT - 2013
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A charge \(q\) is placed at the centre of the line joining two equal positive charges \(Q.\) The system of the three charges will be in equilibrium, if \(q\) is equal to:
| 1. | \(\dfrac{-Q}{4}\) | 2. | \(\dfrac{Q}{4}\) |
| 3. | \(\dfrac{-Q}{2}\) | 4. | \(\dfrac{Q}{2}\) |
Subtopic: Â Coulomb's Law |
 66%
From NCERT
NEET - 2013
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Two positive ions, each carrying a charge \(q\), are separated by a distance \(d\). If \(F\) is the force of repulsion between the ions, the number of electrons missing from each ion will be:
(\(e\) is the charge on an electron)
| 1. | \(\frac{4 \pi \varepsilon_{0} F d^{2}}{e^{2}}\) | 2. | \(\sqrt{\frac{4 \pi \varepsilon_{0} F e^{2}}{d^{2}}}\) |
| 3. | \(\sqrt{\frac{4 \pi \varepsilon_{0} F d^{2}}{e^{2}}}\) | 4. | \(\frac{4 \pi \varepsilon_{0} F d^{2}}{q^{2}}\) |
Subtopic: Â Coulomb's Law |
 78%
From NCERT
AIPMT - 2010
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- 1(current)
Q. No.
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