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Electric Charges And Fields

Question

Two metal plates having a potential difference of 800 V and 2 cm apart. It is found that a particle of mass 1.96 x 10^-15 kg remaining suspended in the region between the plates. The charge on the particle must be (e = elementry charge)

8e

6e

3e

2e

Solution

$E = \frac{V}{d} = \frac{800}{2 \times 10^{- 2}} \frac{V}{m = 4 \times 10^{4} \frac{V}{m}} again, qE = mg q = \frac{mg}{E} = \frac{1.96 \times 10^{- 15} \times 9.8}{4 \times 10^{4}} e q = 3 e$

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Electric Charges And Fields

Two metal plates having a potential difference of 800 V and 2 cm apart. It is found that a particle of mass 1.96 x 10^-15 kg remaining suspended in the region between the plates. The charge on the particle must be (e = elementry charge)

8e

6e

3e

2e

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For Daily Free Study Material Join wiredfaculty

Electric Charges And Fields

Two metal plates having a potential difference of 800 V and 2 cm apart. It is found that a particle of mass 1.96 x 10-15 kg remaining suspended in the region between the plates. The charge on the particle must be (e = elementry charge) 8e 6e 3e 2e

Some More Questions From Electric Charges and Fields Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Two metal plates having a potential difference of 800 V and 2 cm apart. It is found that a particle of mass 1.96 x 10^-15 kg remaining suspended in the region between the plates. The charge on the particle must be (e = elementry charge)

8e

6e

3e

2e