Question

An oil drop of mass 50mg and of charge -5uC is just balanced in air against the force of gravity
[g=9.8m/s2]. Calculate the strength of the electric field required to balance it.
A) 98 N/C upwards
B) 98 N/C downwards
C) 9.8 N/C towards north
D) 9.8 N/C towards south​

Answer 1

Given:

An oil drop of mass 50mg and of charge -5uC is just balanced in air against the force of gravity [g=9.8m/s²].

To find:

Magnitude and direction of electric field?

Calculation:

Let the electrostatic field intensity be E :

• Now, since the oil drop is in equilibrium the electrostatic force is balancing the weight of the oil drop.

• Since weight is downwards, the electrostatic field intensity has to be directed upwards in order to balance it.

$\rm \: F_{electrostatic} = F_{gravity}$

$\rm\implies \: Eq = mg$

• Convert mass from mg to kg.

$\rm\implies \: E \times (5 \times {10}^{ - 6} )= \dfrac{50}{1000000} \times 9.8$

$\rm\implies \: E \times (5 \times {10}^{ - 6} )= \dfrac{5}{100000} \times 9.8$

$\rm\implies \: E \times {10}^{ - 6} = {10}^{ - 5} \times 9.8$

$\rm\implies \: E = 9.8 \times 10$

$\rm\implies \: E = 98 \: N/C$

So, field intensity required is 98 N/C upwards .

Answer 2

Answer:

b

Explanation: