How many electrons should be removed from a conductor so that it acquires a positive charge of 3.5nC?

Open in App

Solution

Given
Charge $q = 3.5 \times 10^{- 9}$

Let the $n Q$ of electron to be removed be $n$
We know that
$Q = n e$

$n = \frac{Q}{e}$

$n = \frac{3.5 \times 10^{- 9}}{1.6 \times 10^{- 19}}$

$n = 2.1875 \times 10^{- 9 + 19}$

$n = 2.1875 \times 10^{10}$

Suggest Corrections

Similar questions

The number of electrons must be removed from a conductor, so that it acquires a charge of 4.8nC

Q. Calculate how many electrons may be removed from a body so that it may have a charge of

+ 1

coulomb

Q. How many electrons would have to be removed from a copper penny to leave it with a positive charge of

10^{- 7} C

Q. The number of electrons to be removed from a glass rod in order that it acquires a charge of

1 μ C

is:

Analyze the given statements and choose the correct option:

Statement1: If a glass rod is rubbed with silk cloth it acquires positive charge.

Statement2: A body acquires a positive charge when electrons are removed from it.

Join BYJU'S Learning Program

How many electrons should be removed from a conductor so that it acquires a positive charge of 3.5nC?

Given Charge q=3.5×10−9Let the nQ of electron to be removed be nWe know that Q=nen=Qen=3.5×10−91.6×10−19n=2.1875×10−9+19n=2.1875×1010

Given
Charge $q = 3.5 \times 10^{- 9}$

Let the $n Q$ of electron to be removed be $n$
We know that
$Q = n e$

$n = \frac{Q}{e}$

$n = \frac{3.5 \times 10^{- 9}}{1.6 \times 10^{- 19}}$

$n = 2.1875 \times 10^{- 9 + 19}$

$n = 2.1875 \times 10^{10}$