Two charges are placed on the x axis, +5.00 μC at the origin and -10.0 μC at x=10.0 cm x=10.0

Two charges are placed on the x axis, +5.00 μC at the origin...

Two charges are placed on the $x$ axis, $+5.00 \mu \mathrm{C}$ at the origin and $-10.0$ $\mu \mathrm{C}$ at $\mathrm{x}=10.0 \mathrm{~cm} x=10.0 \mathrm{~cm}$. (a) Find the electric field on the $x$ axis at $\mathrm{x}=6.00 \mathrm{~cm} x=6.00 \mathrm{~cm}$. (b) At what point(s) on the $x$ axis is the electric field zero? Example 16-6

Key Concepts

Coulomb's Law

Coulomb's Law describes the electric force or field produced by a point charge. It states that the magnitude of the electric field produced by a charge is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance from the charge. This law is fundamental for computing the electric field produced by point charges and applies through the constant k, often expressed as 1/(4???).

Electric Field as a Vector Field

The electric field is a concept that represents the effect that a charge exerts on the space around it, influencing other charges by exerting a force. It is a vector quantity, meaning it has both magnitude and direction. This property requires careful consideration of both the size and the orientation of the field contribution when combining fields from multiple sources.

Superposition Principle

The superposition principle states that the net electric field created by multiple charges is the vector sum of the individual electric fields produced by each charge. This principle is crucial when analyzing systems with more than one charge because it allows us to consider the combined effect by individually calculating and then summing the fields due to each charge.

Electric Field Null Points

Determining points where the electric field is zero involves setting the net electric field equal to zero and solving for the position. In systems with multiple charges, this typically requires balancing the contributions of fields from each charge, taking into account both their magnitudes and directions. The location of such points depends on the relative sizes of the charges and the distances from them, and it often leads to solving equations based on Coulomb's Law.

Transcript

00:01 This problem covers the concept of the electric field produced by a charge and we are going to use this equation to find the net electric field at point p.

00:13 So for part a, the electric field at p by q1 equals the column constant that is 8 .99 into 10 raise 9 newton meter square or column square.

00:35 Into the absolute value of the charge q1 that is 5 microculum or 5 into 10 res minus 6 cool up upon square of the distance that is 0 .06 meter square so the magnitude of the electric field at point p by q1 equals 1 .25 into 10 .7 newton for coolum now the electric field at point p by q2 equals 8 .99 into 10 raise 9 newton meter square per column square into the absolute value of the charge q2 that is 10 microculum or 10 into 10 res minus 6 column upon the square of the distance that is 0 .04 meter square so the magnitude of the electric field at p by q2 equals 5 .62 into 10 .7 newton for kuna.

01:49 Now since q1 and q2 are opposite sign and q1 is negative and q2 is negative and q1 is positive therefore the electric field at point p by q1 is along the positive x direction and also the electric field by q2 on p is also along the electric field by q2 on p is also along the positive x direction.

02:13 So the net electric field at p equals the vector sum of the electric field at p by q1 plus the electric field at p by q2 or we can write the net electric field equals 1 .25 into 10 .7 newton per column plus 5 .62 into 10 .7 newton per column plus 5 .62 into 10 .7 newton per column or the net electric field at p equals 6 .89 into 10 race 7 newton per column and the direction is along positive x -axis.

03:12 Now part p.

03:17 Consider a point to consider point p left to q1 and at a distance of x from q1.

03:28 So we can write the net electric field p equals kq1 upon x square with negative sign because it is pointing towards the negative x -axis plus kq2 upon x2 plus x2 plus x square and that is equivalent to 0.

03:59 So we can cancel out k and k on the both terms...

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