SATHEE: Electrostatics Question 219

Electrostatics Question 219

Question: The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times 10^{- 11} m,$ will be (Charge on electron $= 1.6 \times 1 0^{19} C$ , mass of electron $= 9.1 \times 1 0^{31} k g$ , mass of proton = $1.6 \times 10^{- 27} k g,$

$G = 6.7 \times 10^{- 11} N m^{2} / k g^{2})$ [RPET 1997; Pb PMT 2003]

Options:

A) $2.36 \times 1 0^{39}$

B) $2.36 \times 1 0^{40}$

C) $2.34 \times 1 0^{41}$

D) $2.34 \times 1 0^{42}$

Show Answer

Answer:

Correct Answer: A

Solution:

Gravitational force $F_{G} = \frac{G m_{e} m_{p}}{r^{2}}$

$F_{G} = \frac{6.7 \times 10^{- 11} \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 27}}{{(5 \times 10^{- 11})}^{2}}$ = 3.9x 10?47 N Electrostatic force $F_{e} = \frac{1}{4 π ε_{0}} \frac{e^{2}}{r^{2}}$

$F_{e} = \frac{9 \times 10^{9} \times 1.6 \times 10^{- 19} \times 1.6 \times 10^{- 19}}{{(5 \times 10^{- 11})}^{2}}$ = 9.22x 10?8 N So, $\frac{F_{e}}{F_{G}} = \frac{9.22 \times 10^{- 8}}{3.9 \times 10^{- 47}} = 2.36 \times 10^{39}$

Electrostatics Question 219

Question: The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times 10^{- 11} m,$ will be (Charge on electron $= 1.6 \times 1 0^{19} C$ , mass of electron $= 9.1 \times 1 0^{31} k g$ , mass of proton = $1.6 \times 10^{- 27} k g,$

Options:

Answer:

Solution:

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Electrostatics Question 219

Question: The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance 5×10−11m, will be (Charge on electron = 1.6×1019C , mass of electron = 9.1×1031kg , mass of proton = 1.6×10−27kg,

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: The ratio of electrostatic and gravitational forces acting between electron and proton separated by a distance $5 \times 10^{- 11} m,$ will be (Charge on electron $= 1.6 \times 1 0^{19} C$ , mass of electron $= 9.1 \times 1 0^{31} k g$ , mass of proton = $1.6 \times 10^{- 27} k g,$