In the given figure three point charges are situated

\( \newcommand\) • • • • • • • Overview Recall that the electric potential is defined as the electric potential energy per unit charge \[\mathrm \] Point Charges Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical charge distributions (like on a metal sphere, see figure below) create external electric fields exactly like a point charge. The electric potential due to a point charge is, thus, a case we need to consider. Using calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge Q, and noting the connection between work and potential (W=–qΔV), it can be shown that the electric potential V of a point charge is \(\mathrm \] The electric potential is a scalar while the electric field is a vector. Note the symmetry between electric potential and gravitational potential – both drop off as a function of distance to the first power, while both the electric and gravitational fields drop off as a function of distance to the second power. Superposition of Electric Potential We’ve seen that the electric potential is defined as the amount of potential energy per unit charge a test particle has at a given location in an electric field, i.e. \[\mathrm \right]\] To find the total electric field, you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact that V is closely associated with energy, a sc...

In fig. 1(a).21, let Q be the charge required to be placed at G. Now. − → F 1 = force at A due to charge at B = 1 4 π ∈ 0 q 2 l 2 , along − − → B A − → F 2 = force at A due to charge at C = ( 1 4 π ∈ 0 q 2 l 2 ), along − − → C A. − → F 1 + − → F 2 = √ F 2 1 + F 2 2 + 2 F 2 F 2 cos 60 0 = F 1 √ 3 = √ 3 q 2 4 π ∈ 0 l 2 along GA. Force at A due to charge Q at the centroid G = 1 4 π ∈ 0 Q q A G 2 = Q q 4 π ∈ 0 ( l / √ 3 ) 2 = 3 Q q 4 π ∈ 0 l 2 This must be equal and opposite to ( − → F 1 + − → F 2 ) ∴ 3 Q q 4 π ∈ 0 l 2 = − √ 3 q 2 4 π ∈ 0 l 2 Q = q √ 3

• • • • Question:Three point charges are located on the corners of a right triangle as shown in the Figure, given: What is the magnitude (unit in N ) of the electric force of q2 onto q3 ? Question 10 Three point charges are located on the corners of a right triangle as shown in the Figure, given: What is the x-component (unit in N ) of the electric force of q1 onto q3 ? Three point charges are located on the corners of a right triangle as shown in the Figure, given: What is the magnitude (unit in N ) of the electric force of q 2 ​ onto q 3 ​ ? Question 10 Three point charges are located on the corners of a right triangle as shown in the Figure, given: What is the x-component (unit in N ) of the electric force of q 1 ​ onto q 3 ​ ? Previous question Next question

Consider three point charges located at the corners of a right triangle as shown in Figure below, where Coulomb force: We know that force on any charge due to the vicinity of another charge is given by the Coulomb force. According to Coulomb's law, the force is directly proportional to the product of the charge and inversely proportional to the square of the distance between them. Answer and Explanation: 1

from CORRECTION: At 24.33 min multiply the expression by sin (theta). Example 2- Three Point Charges Now, let’s consider another example. In this case, let’s assume that we have three point charges, which are located at the corners of a right triangle. Let’s say we have a positive charge, q1 is located at the top corner, and negative charge – q2 is located at the lower left hand corner, and another negative, – q3 is located on the right hand corner of this right triangle. This also gives some dimensions to the distances. Say, the distance between q1 and q2 is a, q2 and q3 is b, and what we’re interested in is the net force on charge q3 due to the charges q1 and q2 is the question mark. Alright. The first step will be determining the directions of the forces generated on q3 due to q1 and q2 by considering the sign of the pair of these charges. Well, q1 is positive, q3 is negative. Therefore it will attract q3 along the line which joins these two charges. Let’s denote this force as F31, which is the force on q3 due to q1. Similarly, when we look at the force generated on q3 by q2, since both of them are like charges, therefore q2 will repel q3 with a force of F32, and this is force on q3 due to q1. One we determine the directions of the forces generated by q1 and q2 on q3, the next step just becomes the vector addition of these two forces to be able to get the total force acting on charge q3. In order to do that, by recalling the vector addition rule, which states that vecto...