The electric field is created at point P 1 by charge Q 1 has a magnitude of: The

Question

The electric field is created at point P1 by charge Q1 has a magnitude of:

The electric field created at point P1 by charge Q1 is directed:

The total electric field at point P1 is equal to:

The electric field created at point P2 by charge Q1 is directed:

The electric field created at point P2 by charge Q2 has a magnitude of:

The electric field created at point P2 by charge Q2 is directed:

The scalar x component of the total electric field at point P2 equals:

The scalar y component of the total electric field at point P2 equals:

If a third charge Q3 = -2.00 mircocoulombs were placed at point P2, Q3 would experience an electostatic force of magnitude:

Please SHOW ALL WORK! Thank you!

The figure below shows two point particles having charges Q1-Q and Q2 = + Q , where Q = 0.100 C . Suppose that the two charged particles are held fixed in position. Q=0.100 pC "P1" and "P2" are just points in the x-y plane they are not charged particles. 8.00 cm- The point “P1" is located midway between the two charged particles. P1 +X 3.00 cm P2

Explanation / Answer

Solution:

The electric field created by Q1 at the point P1 is along the negative X axis, since Q1 is negative.

1) Its magnitude is given by E1=kQ1/ (0.04)2

wheere k=9 e9 Nm^2/C^2 is the Coulomb's constant and Q1 = -0.1 uC = 0.1 x 10^-6 C

E1= (9 e9)(-1 e-7) / (0.04)2 = - 5.63 e5 N /C

2 ) The direction of the electric field created by P1 is directed along the negative X axis, since the charge Q1 is Negative.

3 ) The total electric field at P1 = E1+ E2

Since Q1=Q2 = 0.1 * 10^-6 C, and distance to P1 is 0.04 m from each charge,

E1 = E2 = 5.63 x10^5 N/C , in magnitude;

Direction of E1 is to the left and E2 is also to the left;

The total electric field at P1 E = (-5.63 x 10^5) + (-5.63 x10^5) = -1.3 x 10^6 N/C

4)

Distance to P2 from Q1 = square root [(0.04)^2 + (0.03)^2 ] = 0.05 m

Angle madeby the point P2 with X axis is tan^-1(3/4) = 36.9 degrees

The electric field created at P2 = E2 = kQ1/(0.05)2 = (9 e9)(-.1 * e-6) / (0.05)2 = -3.6 x 105 N/C

The X component of E2lectric field at P2 by Q2 = E2x = E2 cos 36.9 = (-3.6 x 10^5) cos 36.9 =- 2.88 x 10^5 N/C

The Y component of E2 = E2 sin36.9 = (3.6 x 10^-5)(sin36.9) = -2.16 x 10^5 N/C

5) Direction of E2 = tan^-1 (E2y/E2x) = 36.9 degree

5)

The scalar x component of the total electric field at point P2 equals = 2.88 x 10^5 N/C

The scalar y component of the total electric field at point P2 equals = 2.16 x 10^5 N/C

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