Electric Charges And Fields
Electrostatic force
F= kq1q2/r2
As q1=q2=1.6 * 10-19 for alpha particles
R=3.2 * 10-15
And K=9 * 109
Therefore, F= (9 * 109) * (1.6 * 10-19) * (1.6 * 10-19)/ (3.2 * 10-15)2
è F= (23.04 * 10-29)/(10.24 * 10-30)
è F=2.25 * 10
è F=22.5 N
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QUESTION:- Calculate the coulomb force between two Alpha particles separated by a distance of 3.2X10 to the power -15 m in air.
ams 92.16x10^-3 and 2.97x10^-37
This will help you
Four point charges qA = 2 μC, qB = – 5 μC, qc = 2 μC, and qD = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Suppose a square ABCD with each side of 10 cm and centre 0. At the centre, the charge of 1 μC is placed.
As qA = qc, the charge of 1 μC experiences equal and opposite forces FA and Fc due to charges qA and qC.
At the same time, the charge 1 μC experiences equal and opposite forces. FB and FD due to equal charges qB and qD at B and D. Thus, the net force on charge of 1 μC due to the given charges is zero.
Text Solution
`90 N``45 N``60 N``75 N`
Answer : A
Solution : Here, `r= 3.2xx10^(-15)m` <br> charge on `alpha`-particle, `q_(1)=q_(2)= 2xx1.6xx10^(-19)C` <br> Now, `F= 1/(4piepsilon_(0))*(q_(1)q_(2))/(r^(2))` <br> `=9 xx 10^(9)xx((2xx1.6xx10^(-19))^(2))/((3.2xx10^(-15))^(2))=90 N`