A charged particle P has a mass of `10^(-16)`kg and carries a charge of `4.9 xx 10^(-19)C`. Calculate the intensity of the electric field to be applied on it in vertically upward direction, so as to keep it at rest.

Answer

To solve the problem of finding the intensity of the electric field required to keep a charged particle at rest, we can follow these steps:

### Step 1: Identify the Forces Acting on the Particle
The charged particle experiences two main forces:
1. The gravitational force acting downward, which is given by \( F_g = mg \).
2. The electric force acting upward due to the electric field, which is given by \( F_e = QE \).

### Step 2: Set Up the Equation for Equilibrium
For the particle to remain at rest, the upward electric force must balance the downward gravitational force. Therefore, we can set up the equation:
\[ QE = mg \]

### Step 3: Rearrange the Equation to Solve for Electric Field \( E \)
We can rearrange the equation to solve for the electric field \( E \):
\[ E = \frac{mg}{Q} \]

### Step 4: Substitute the Given Values
Now, we substitute the given values into the equation:
- Mass \( m = 10^{-16} \) kg
- Charge \( Q = 4.9 \times 10^{-19} \) C
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Substituting these values into the equation gives:
\[ E = \frac{(10^{-16} \, \text{kg})(10 \, \text{m/s}^2)}{4.9 \times 10^{-19} \, \text{C}} \]

### Step 5: Calculate the Electric Field \( E \)
Now, we perform the calculation:
\[ E = \frac{10^{-15}}{4.9 \times 10^{-19}} = \frac{10^{-15}}{4.9} \times 10^{19} \]
\[ E \approx 2.04 \times 10^{3} \, \text{N/C} \]

### Final Answer
The intensity of the electric field required to keep the charged particle at rest is approximately:
\[ E \approx 2.04 \times 10^{3} \, \text{N/C} \]

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