Home › UP Board Class 12 › Physics › Electric Charges and Fields › A point charge of $5\ \mu$C is placed at the ori…
A point charge of $5\ \mu$C is placed at the origin. The electric field magnitude at $0.5$ m is:
A$1.8\times 10^5$ N/C, by $E = kQ/r^2$
B$9\times 10^4$ N/C, half the correct value here
C$0$ N/C, since the point is far on the table
D$5\times 10^{-6}$ N/C, the charge in coulombs
Answer & Solution
Correct answer: A. $1.8\times 10^5$ N/C, by $E = kQ/r^2$
$E = 9\times 10^9\cdot 5\times 10^{-6}/0.25 = 1.8\times 10^5$ N/C.
Related questions
A charged comb picks up small pieces of paper that carry no net charge, because the paper Compared with the tiny nucleus, the electrons around it form a cloud that is:Benjamin Franklin explained early observations by supposing that one type of charge stayedA test charge or a source charge moves, and as a result the force between them:If the two ends of a dipole can be pulled apart, induction can create charged objects withPlaced in a uniform field, such a dipole experiences a turning effect that changes its angTwo equal and opposite charges separated by a small distance form a dipole, characterised Where those lines are packed more densely, the field is: