SATHEE: Electrostatics Question 208

Electrostatics Question 208

Question: An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force $\vec{F}$ between the two is (Where $K = \frac{1}{4 π ε_{0}}$ ) [CBSE PMT 2003]

Options:

A) $- K \frac{e^{2}}{r^{3}} \hat{r}$

B) $K \frac{e^{2}}{r^{3}} \vec{r}$

C) $- K \frac{e^{2}}{r^{3}} \vec{r}$

D) $K \frac{e^{2}}{r^{2}} \hat{r}$

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Answer:

Correct Answer: C

Solution:

$\vec{F} = - k \frac{e^{2}}{r^{2}} \hat{r} = - k . \frac{e^{2}}{r^{3}} \vec{r}$

$(∵ \hat{r} = \frac{\vec{r}}{r})$

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Electrostatics Question 208

Question: An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force F→ between the two is (Where K=14πε0 ) [CBSE PMT 2003]

Options:

Answer:

Solution:

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Question: An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The coulomb force $\vec{F}$ between the two is (Where $K = \frac{1}{4 π ε_{0}}$ ) [CBSE PMT 2003]