For the two point charges Q1 and Q2 shown below, findthe electric potential VA at point A,the electric potential VB at point B, andthe electric potential difference ΔV=VB - VA The image depicts a diagram illustrating the positions of two charges and a point of interest, labeled as A and B, with specified distances. Here's a detailed description:- **Points and Charges:** - Point **A** is 30 cm above the x-axis. - Point **B** is on the y-axis, 60 cm from point A. - **Q₂ (at the bottom left)** is a positive charge of +50 µC located 30 cm directly below point A on the y-axis. - **Q₁ (at the bottom right)** is a negative charge of -50 µC positioned 26 cm to the right along the x-axis.- **Distances:** - The direct horizontal distance between Q₂ and the origin (intersection of x and y axes) is 26 cm. - The direct horizontal distance between Q₁ and the origin is also 26 cm. - The distance between Q₂ and Q₁ forms a right triangle, with Q₂-Q₁ as the hypotenuse, marked as 40 cm.This diagram can be used to analyze electric fields, forces between charges, or potential calculations in the context of electrostatics. The positioning of the charges and the defined distances are critical for solving related physics problems using formulas for electric force or field.

Answered: B В A 60 cm 30 cm !! 26 cm 26 сm Q1…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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For the two point charges Q₁and Q₂shown below, find

the electric potential V_A at point A,
the electric potential V_B at point B, and
the electric potential difference ΔV=V_B - V_A

The image depicts a diagram illustrating the positions of two charges and a point of interest, labeled as A and B, with specified distances. Here's a detailed description:

- **Points and Charges:**
- Point **A** is 30 cm above the x-axis.
- Point **B** is on the y-axis, 60 cm from point A.
- **Q₂ (at the bottom left)** is a positive charge of +50 µC located 30 cm directly below point A on the y-axis.
- **Q₁ (at the bottom right)** is a negative charge of -50 µC positioned 26 cm to the right along the x-axis.

- **Distances:**
- The direct horizontal distance between Q₂ and the origin (intersection of x and y axes) is 26 cm.
- The direct horizontal distance between Q₁ and the origin is also 26 cm.
- The distance between Q₂ and Q₁ forms a right triangle, with Q₂-Q₁ as the hypotenuse, marked as 40 cm.

This diagram can be used to analyze electric fields, forces between charges, or potential calculations in the context of electrostatics. The positioning of the charges and the defined distances are critical for solving related physics problems using formulas for electric force or field.

Expert Solution

Step 1 - Electric potential at A

Given :

Q1 = -50 $μ C = - 50 * 10^{- 6} C$

Q2 = 50 $μ C = 50 * 10^{- 6} C$

Distance between A and Q1 = 60 cm = 0.6 m

Distance between A and Q2 = 30 cm = 0.3 m

Distance between B and Q1 = 40 cm = 0.4 m

Distance between A and Q2 = 40 cm = 0.4 m

By the formula of Electric potential due to point charge :

$V = \frac{1}{4 {πε}_{°}} . \frac{Q}{r}$ ; where Q = magnitude of charge and r = distance of point from the charge and

$\frac{1}{4 {πε}_{°}} = 9 * 10^{9}$

Applying this formula ; to find potential at point A

$V_{A} = \frac{1}{4 {πε}_{°}} . \frac{Q_{1}}{D i s \tan c e f r o m Q 1 t o A} + \frac{1}{4 {πε}_{°}} . \frac{Q_{2}}{D i s \tan c e f r o m Q 2 t o A}$

Substituting the given values :

$V_{A} = 9 * 10^{9} * (\frac{- 50 * 10^{- 6}}{0.6} + \frac{50 * 10^{- 6}}{0.3})$

On solving, we get:

$V_{A} = 74.97 * 10^{4} V$

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