The force on a unit positive charge at the given position is determined as 1.8 x 10⁻⁸ N.
<h3>Force on the charge</h3>
The force on the charge is calculated as follows;
F = k(q1q2)/r²
where;
- r is the distance between the charges = 2 m - 1m = 1m
F = (9 x 10⁹ x 10⁻⁹ x 2 x 10⁻⁹)/(1)²
F = 1.8 x 10⁻⁸ N
Thus, the force on a unit positive charge at the given position is determined as 1.8 x 10⁻⁸ N.
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Answer:
A linear, positive relationship
Explanation:
The graph is a line, so it is linear and the slope is positive, so it is positive...
Answer:
B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.
Explanation:
Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.
Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.
So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.
Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.
Also,option D is ruled out as ammeter is connected in series.
