Resistance = (voltage) / (current)
Resistance = (12v) / (0.33 A)
Resistance = (12/0.33) ohms
<em>Resistance = 36.4 ohms</em>
The resultant force on the positive charge is mathematically given as
X=40N
<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>
Question Parameters:
Three-point charges, two positive and one negative, each having a magnitude of 20
Generally, the -ve charge is mathematically given as

Q+=X
Therefore

X=40N
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Solution:
Let the slope of the best fit line be represented by '
'
and the slope of the worst fit line be represented by '
'
Given that:
= 1.35 m/s
= 1.29 m/s
Then the uncertainity in the slope of the line is given by the formula:
(1)
Substituting values in eqn (1), we get
= 0.03 m/s
Answer:
Electric potential = 0.00054 V
Explanation:
We are given;
Charge; q = 3 pC = 3 × 10^(-12) C
Radius; r = 2 cm = 0.02 m
Formula for the electric potential of this surface will be;
V = kqr
Where;
K is a constant = 9 × 10^(9) N⋅m²/C².
Thus;
V = 9 × 10^(9) × 3 × 10^(-12) × 0.02
V = 0.00054 V