ou quick no longer be able to you have on the bookmarks bar, lmport bookmarks no

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ou quick no longer be able to you have on the bookmarks bar, lmport bookmarks now... 15403836 purchased access online. Th Q4 an access code for this class or or redeemed assignments not yet your WebAssign here see Chegg.com bookmarks records your to www.webassign.net/web/student/Assignment-Responses/submit?dep place our access, D According grades Get access now. O-1 points serEssen 15 AE 04 Example 15.4 Electric Field Due to Two Point Charges Goal Use the superposition principle to calculate the electric field due to two point charges. Problem Charge q1 6.65 HC is at the origin, and charge q2 -5.05 uC is on the x-axis, 0.300 m from the origin (Fig. 15.12) (a) Find the magnitude and direction of the electric field point P, which has coordinates (0, 0.400) m. (b) Find the force on a charge of 2.00 x 10 8 c placed at P. Figure 15.12 e resultant electric field E at p equals the vector sum E1 E2, where E1 is the field due to the positive charge q1 and 2 is the field due to the negative charge g2 strategy the problem the electric field at point P due to each individual in terms of x- components, then adding the components of each type to get the components of the resultant electric field at P. he magnitude of the force in part (b) can simply multiplying the magnitude of the electric field be found by by the charge. (a) Calculate the electric field at P. Find the magnitude of E 1 with Equation 15. 6.63 x 10 c (8.99 x 10 N m2/cz) (0.100 m) 01w

Explanation / Answer

Electric filed at the location of q2 due to -10.60 C at P = (9x109)(10.60x10-6)/(0.500)2 = 0.3816x106 N/C towards P

X component of this field = - (0.3816x106) cos (theta) = - 0.22896x106 N/C or 0.22896x106 N/C along – ve x - axis.

Y component of this field = (0.3816x106) sin (theta) = 0.30528x106 N/C along + y axis.

Electric filed at the location of q2 due to q1(6.65 C) = (9x109)(6.65x10-6)/(0.300)2 = 0.665x106 N/C along + x axis.

Hence

X component of net electric field = 0.43604x106 N/C

Y component of net electric field = 0.30528x106 N/C

And

Magnitude of net electric field = 0.5322844728x106 N/C

Direction = 34.99660572 degree with x – axis.

Also the magnitude of force on q2 = (q2)(E) = (5.05x10-6)(0.5322844728x106)= 2.688036588 N

Direction of force = 214.99660572 degree with x – axis.

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