In Millikan's experiment,an oil drop having c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For the oil drop to be stationary,the electric force acting on it must balance its weight.The electric force $F_e$ on a charge $q$ in an electric

Vedclass · Accepted Answer

$q\frac{V}{d}$

Vedclass · Answer

(D) For the oil drop to be stationary,the electric force acting on it must balance its weight.The electric force $F_e$ on a charge $q$ in an electric field $E$ is given by $F_e = qE$.The electric field $E$ between two plates separated by a distance $d$ with a potential difference $V$ is given by $E = \frac{V}{d}$.Equating the electric force to the weight of the drop $(mg)$:$mg = qE$$mg = q\left(\frac{V}{d}\right)$Therefore,the weight of the drop is $q\frac{V}{d}$.

In Millikan's experiment,an oil drop having charge $q$ becomes stationary upon applying a potential difference $V$ between two plates separated by a distance $d$. The weight of the drop is:

Similar Questions

Seven balls are drawn simultaneously from a bag containing $5$ white and $6$ green balls. The probability of drawing $3$ white and $4$ green balls is:

The $P-V$ diagram of $2\,g$ of helium gas for a certain process $A \to B$ is shown in the figure. What is the heat given to the gas during the process $A \to B$ (in $,P_0V_0$)?

The major product of the reaction is:

$A$ roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20 \, m$. The speed of the car at the top of the hill is between

Which of the following is considered a hotspot of biodiversity in India?

Vedclass Test Series

Exam Paper Generator

Online Exam Module