Please I need some helptI need help please Thank you The fig

Please I need some helptI need help please. Thank you

The figure below shows two point particles having charges Q_1 = - Q and Q_2 = +Q, where Q = 0.100 mu C. Suppose that the two charged particles are held fixed in position. \"P1\" and \"P2\" are just points in the x-y plane, they are not charged particles. The point \"P_1\" is located midway between the two charged particles. The electric field created at point P_1 by charge Q_1 has a magnitude of 5.62 Times 10^1 N/C. 00562 N 2.25 Times 10^5 N/C 5.62 Times 10^5 N/C 2.25 Times 10^4 N/C The electric field created at point P_1 by charge Q_1 is directed in the negative y direction in the positive y direction. in the negative x direction in the positive x direction none of the above The total electric field at point P_1 is equal to zero 1.12 Times 10^7 N/C in the negative Times direction 1.12 Times 10^2 N/C in the positive x direction. 4.50 Times 10^4 N/C in the negative x direction 1.12 Times 10^6 N/C in the positive Times direction in the negative x direction The electric field created at point P_2 by charge Q_1 is directed 36.9 degrees above the negative Times direction 53.1 degrees below the positive Times direction, 36.9 degrees below the negative Times direction. 53.1 degrees above the positive Times direction 53.1 degrees below the negative x direction The electric field created at point P_12 by charge Q_2 has a magnitude of 1.83 Times10^5 N/C. 1.23 Times 10^5 N/C. 5.62 Times 10^5 N/C. 3.60 Times 10^5 N/C. 56.2 N/C. The electric field created at point P_2 by charge Q_2 is directed 36.9 degrees above the negative Times direction. 36.9 degrees below the positive Times direction 36.9 degrees below the negative Times direction. 36 9 degrees above the positive Times direction 53.1 degrees below the negative Times direction The scalar x component of the total electric field at point P_2 equals - 2.88 Times 10^5 N/C. 2.16 Times10^5 N/C. - 4.32 Times 10^5 N/C. -5.75 Times 10^5 N/C. zero The scalar y component of the total electric field at point P_2 equals - 2.70 Times 10^5iN/C. 5.40 Times 10^5 1.20 Times 1.20 Times 10^5 N/C. - 5.75 Times 10^5 N/C. zero If a third charge Q_3 = - 2.00 micro-coulombs were placed at point P_2, directed Q_3 would experience an electrostatic force c. in the positive y direction equal to zero. in the negative y direction. in the negative x direction, in the positive x direction. The force would be

Solution

1)

E1 = k*Q1/r1^2

= 9*10^9*(0.1*10^-6)/0.04^2

= 5.62*10^5 N/C <---- option d

2)

Direction = in negative x direction <---- option c

As the charge is negative to electric field is dirceted towards left

3)

E1 = 5.62*10^6 N/C towards left

E2 = k*Q2/r2^2

= 9*10^9*(0.1*10^6)/0.04^2

= 5.62*10^5 N/C towards left

So, Enet = E1 + E2 = 2*5.62*10^5

= 1.12*10^6 N/C in the negative x direction

8)

E = k*Q2/d^2

= 9*10^9*(0.1*10^-6)/0.05^2

= 3.6*10^5 N/C <---- option d

7)

angle, = atan(3/4) = 36.9 deg above negative x axis

9)

angle = atan(3/4) = 36.9 deg below the negative x axis

10)

Ex =- 2*3.6*10^5*cos(36.9 deg)

= -5.75*10^5 N/C

11)

y component = 0 <---- option e

IT is due to the fact that the y components cancel each other out

12)

Enet is in -ve axis

As Q3 = is negative

So, Force will be directed in positive x direction <---- option d