tebogomashiane0
Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

two particles woth each charge magnitude 2.0×10^-7 c but opposite signs are held 15cm apart.what are the magnitude and direction of the electric field E at tge point midway between charges​

Sagot :

onyebuchinnaji

Answer:

The magnitude of the electric field strength is 6.4 x 10⁵ N/C, directed from positive particle to negative particle.

Explanation:

Given;

charge of each particle, Q = 2 x 10⁻⁷ C

distance between the two charges, r = 15 cm = 0.15 m

distance midway between the charges = 0.075 m

The magnitude of the electric field is calculated as;

[tex]E_{net} = E_{+q} + E_{-q}\\\\E_{net} = \frac{kQ}{r_{1/2}^2} + \frac{kQ}{r_{1/2}^2}\\\\E_{net} = 2(\frac{kQ}{r_{1/2}^2})\\\\E_{net} = 2 (\frac{9\times 10^9 \ \times 2\times 10^{-7}}{0.075^2} )\\\\E_{net} = 6.4\times 10^5 \ N/C[/tex]

The direction of the electric field is from positive particle to negative particle.

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.