A -3.00 nC point charge is on the x -axis at x=1.20 m. A second point charge, Q, is on the x -a

A -3.00 nC point charge is on the x -axis at x=1.20 m. A...

A -3.00 nC point charge is on the $x$ -axis at $x=1.20 \mathrm{~m}$. A second point charge, $Q,$ is on the $x$ -axis at $-0.600 \mathrm{~m}$. What must be the sign and magnitude of $Q$ for the resultant electric field at the origin to be (a) $45.0 \mathrm{~N} / \mathrm{C}$ in the $+x$ -direction, (b) $45.0 \mathrm{~N} / \mathrm{C}$ in the $-x$ -direction?

Key Concepts

Coulomb's Law

Coulomb's Law quantifies the electric force between two point charges. It states that the magnitude of the force between two point charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them, with the constant of proportionality being the Coulomb constant (k). This law is foundational in calculating fields and forces in electrostatics.

Electric Field of a Point Charge

The electric field produced by a point charge is given by the expression E = k|q|/r², where q is the charge, r is the distance from the charge, and k is the Coulomb constant. The field points radially outward from a positive charge and radially inward toward a negative charge. This relation is central in determining the influence that a point charge exerts in the space around it.

Superposition Principle in Electrostatics

The superposition principle states that when multiple electric fields are present, the total electric field at any point is the vector sum of the individual fields produced by each charge independently. This principle allows the analysis of complex charge systems by summing the effects of single charges, making it a critical concept in solving electrostatic problems.

Vector Addition

Electric fields are vector quantities, meaning they have both magnitude and direction. To determine the net electric field when multiple fields are present, vector addition is used. This involves both the summing of magnitudes and taking proper account of their directions, which is essential when calculating resultant fields at specific points in space.

Charge Sign and Field Direction

The sign of a charge determines the direction of the electric field it produces: positive charges generate fields that point away from them, while negative charges create fields that point toward them. Understanding this concept is key when setting up problems involving multiple charges, as the combination of their fields depends on both magnitude and directional orientation.

Transcript

00:08 We have small q equals to minus 3 nanowulum, that is 3 into 10 to the power minus 9 koolm and r1 is equal to 1 .20 meter, okay? and second point charge q is on the x -axis which is at a distance r2 equals to 0 .6 0 meter .0 .meter.

00:30 So for the part a, the electric field at the origin due to net electric field from the two charges, small q and capital q, can be written as.

00:38 E small q will be equal to k q by small r1 square so substituting the values k is 9 into 10 to the power 9 and q is minus 3 into 10 to the power 9 so taking the magnitude okay and r1 is 1 .20 square okay so from here eq comes out to be 18 .75 newton per column okay now as the direction of of net electric field in the positive x -axis, this means the net electric field at the origin is given by e equals to eq plus e capital q.

01:18 Okay, so from here we get e capital q equals to e minus e small q.

01:24 So from here we get eq equals to e is given as 45 .0 nanotein per column and e small q is obtained as 18 .75.

01:34 So from here e capital q comes out to be 26 .2.

01:38 Newton per cull.

01:42 We can determine the value of q from the relation kq divided by r2 square and eq is 26 .25.

01:53 So now substituting the values we get q equals to 26 .25 multiplied by r2 is 0 .6 square divided by k is 9.

02:04 Into 10 to the power 9.

02:06 So from here after solving we get capital q equals to 1 .05 nanocolon.

02:15 So this is the answer for this question and this charge will be positive.

02:21 This charge will be positive because the electric field is in the positive direction due to this...

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