Answer :
To determine the intensity of an electric field at a point where a charge experiences a force, we need to use the relationship between the force experienced by a charge in an electric field and the intensity of the electric field.
Here are the steps to solve this problem:
### Step 1: Identifying the Given Data
- Charge ([tex]\(Q\)[/tex]): 0.50 Coulombs (C)
- Force ([tex]\(F\)[/tex]): 20.0 Newtons (N)
### Step 2: Identifying the Unknown
- Electric field intensity ([tex]\(E\)[/tex]): This is what we need to find.
### Step 3: Relevant Equation
The formula that relates the electric field intensity ([tex]\(E\)[/tex]), the force ([tex]\(F\)[/tex]), and the charge ([tex]\(Q\)[/tex]) is:
[tex]\[ E = \frac{F}{Q} \][/tex]
### Step 4: Calculation
Now substitute the given values into the formula:
[tex]\[ E = \frac{F}{Q} = \frac{20.0 \, \text{N}}{0.50 \, \text{C}} \][/tex]
### Step 5: Simplification
Perform the division:
[tex]\[ E = \frac{20.0}{0.50} = 40.0 \, \text{N/C} \][/tex]
### Step 6: Units and Significant Digits
- The unit of electric field intensity is Newtons per Coulomb (N/C).
- The given values (20.0 N and 0.50 C) both have three significant digits, so the result should also have three significant digits.
### Final Answer
The intensity of the electric field is [tex]\( 40.0 \, \text{N/C} \)[/tex].
### Summary
- Given:
- Charge [tex]\(Q = 0.50 \, \text{C} \)[/tex]
- Force [tex]\(F = 20.0 \, \text{N} \)[/tex]
- Equation: [tex]\( E = \frac{F}{Q} \)[/tex]
- Calculation: [tex]\( E = \frac{20.0 \, \text{N}}{0.50 \, \text{C}} = 40.0 \, \text{N/C} \)[/tex]
- Result: [tex]\( \text{Electric field intensity} = 40.0 \, \text{N/C} \)[/tex]
Thus, the intensity of the electric field at the specified point is [tex]\( 40.0 \, \text{N/C} \)[/tex].
Here are the steps to solve this problem:
### Step 1: Identifying the Given Data
- Charge ([tex]\(Q\)[/tex]): 0.50 Coulombs (C)
- Force ([tex]\(F\)[/tex]): 20.0 Newtons (N)
### Step 2: Identifying the Unknown
- Electric field intensity ([tex]\(E\)[/tex]): This is what we need to find.
### Step 3: Relevant Equation
The formula that relates the electric field intensity ([tex]\(E\)[/tex]), the force ([tex]\(F\)[/tex]), and the charge ([tex]\(Q\)[/tex]) is:
[tex]\[ E = \frac{F}{Q} \][/tex]
### Step 4: Calculation
Now substitute the given values into the formula:
[tex]\[ E = \frac{F}{Q} = \frac{20.0 \, \text{N}}{0.50 \, \text{C}} \][/tex]
### Step 5: Simplification
Perform the division:
[tex]\[ E = \frac{20.0}{0.50} = 40.0 \, \text{N/C} \][/tex]
### Step 6: Units and Significant Digits
- The unit of electric field intensity is Newtons per Coulomb (N/C).
- The given values (20.0 N and 0.50 C) both have three significant digits, so the result should also have three significant digits.
### Final Answer
The intensity of the electric field is [tex]\( 40.0 \, \text{N/C} \)[/tex].
### Summary
- Given:
- Charge [tex]\(Q = 0.50 \, \text{C} \)[/tex]
- Force [tex]\(F = 20.0 \, \text{N} \)[/tex]
- Equation: [tex]\( E = \frac{F}{Q} \)[/tex]
- Calculation: [tex]\( E = \frac{20.0 \, \text{N}}{0.50 \, \text{C}} = 40.0 \, \text{N/C} \)[/tex]
- Result: [tex]\( \text{Electric field intensity} = 40.0 \, \text{N/C} \)[/tex]
Thus, the intensity of the electric field at the specified point is [tex]\( 40.0 \, \text{N/C} \)[/tex].
