Electric Field Strength problem.

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SUMMARY

The discussion focuses on calculating the electric field strength at a point X due to two point charges, A (+8μC) and B (-25μC), located 45nm apart in a vacuum. The electric field strength at point X due to charge A is calculated to be 1.2 x 109 N/C, while the overall electric field strength due to both charges is determined to be 1.32 x 109 N/C. The calculations utilize the formula E = (1/4πε) * (Q/d2), highlighting the importance of vector addition when combining electric fields from multiple charges.

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  • Understanding of Coulomb's Law and electric field calculations
  • Familiarity with the concept of point charges
  • Knowledge of vector addition in physics
  • Basic proficiency in using scientific notation
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Darth Frodo
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Homework Statement


2 point charges are situated 45nm apart in a vacuum. Charge A is +8μC and charge B is -25μC. X is a point 8nm from A.

Calculate the electric field strength at point X only due to A (Ans is 1.2 x 10^9 N/C)
Calculate the overall electric field strength at point X due to A and B. (Ans is 1.32 x 10^9 N/C)

Homework Equations


E = [itex]\frac{1}{4\pi\epsilon}[/itex] X [itex]\frac{Q}{d^{2}}[/itex]


The Attempt at a Solution


Electric field strength due to A.
[itex]\frac{1}{4\pi(8.9X10^-12)}[/itex] X [itex]\frac{8X10^-6}{(8X10^-9)^{2}}[/itex]

= 1.117X10^21
= 1.12X10^21 N/C


Electric Field Strength due to B.
E = [itex]\frac{1}{4\pi(8.9X10^-12)}[/itex] X [itex]\frac{25X10^-6}{(37X10^9)^{2}}[/itex]

= 1.632X10^20
= 1.63X10^20 N/C




1.12X10^21 + 1.63X0^20 = 1.283X10^21 N/C


Where am I going wrong?
I would appreciate any help.
 
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It doesn't look like you've done anything wrong. The "answers" given seem to be well out of whack in terms of order of magnitude.
 
I don't see anything wrong with a and b either but the combined field is a vector sum and one of the charges is negative and the other, positive.
 

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