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Chapter 4: Q30E (page 715)

(a) Calculate the magnitude and direction (relative to the +x-axis) of the electric field in Example 21.6. (b) A -2.5-nC point charge is placed at point P in Fig. 21.19. Find the magnitude and direction of (i) the force that the -8.0-nC charge at the origin exerts on this charge and (ii) the force that this charge exerts on the -8.0-nC charge at the origin.

Short Answer

Expert verified
  1. The magnitude and the electric field is 18N and the direction is 127°
  2. (i) Force isF=4.5×10-8N and towards 307°.

(ii) Force is F=4.5×10-8Nand towards 127°.

Step by step solution

01

Step 1:

The magnitude of the electric field is

E=kqr2=kqx2+y22=9×109×8×10-91.22+1.622=18N/C

And the direction of the electric field is

θ=arctan-1.61.2=-53°asitis2ndquadθ=-53+180=127°

Hence, the magnitude and the electric field is 18N and the direction is127°.

02

Step 2:

(i)As the charge at (p) is negative therefore the force is in the opposite direction

F=Eq=18×2.5×10-9=4.5×10-8N

As it is in the forth quad

θf=127+180=307°

Therefore, Force is F=4.5×10-8Nand towards307° .

(ii)The force exerted on charge at the origin is equal to F

role="math" localid="1668174608326" F=4.5×10-8Nangleθ;θ=370-180=127°

Hence, Force is F=4.5×10-8Nand towards 127°.

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