Question

A water particle of mass 10.0 mg and with a charge of 1.50 × 10−6 C stays suspended in a room. What is the magnitude of electric field in the room? What is its direction?

Answer 1

The magnitude of electric field in the room is, E = 65.33 N/C.

Its direction will be upward to balance the downward gravitational force.

Explanation:    

Step 1:

Given data in the question :  

      A particle of water with mass = $10.0 \mathrm {mg}=10 \times 10^{-6} \mathrm{kg}$

                                      $\text { Charge }(q)=1.50 \times 10^{-6} \mathrm{C}$

Step 2:

• "g" is defined as Gravity Acceleration.
• Its Earth value is 9.8 m /s².
• In other words, the earth's surface gravity acceleration at sea level is 9.8 m/s².

                                      F = mg = qE

                                            mg = qE

                                               $E=\frac{m g}{q}$          

                                               $E=\frac{10 \times 10^{-6} \times 9.8}{1.50 \times 10^{-6}}$

                                               $E=\frac{98 \times 10^{-6}}{1.50 \times 10^{-6}}$                    

                                               $E=65.33 \mathrm{N} / \mathrm{C}$    

Thus, the electric field, E = 65.33 N/C and its direction will be upward.

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

63.55 N/C

hope it help