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# Three charges are arranged as shown in Figure P15.11. Find the magnitude and directionof the electrostatic force on the charge at the origin.

## $R=1.38 \times 10^{-5} N,$ at $77.5^{\circ}$ below the $(-\mathrm{x})$ axis.

### Discussion

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##### Christina K.

Rutgers, The State University of New Jersey

##### Andy C.

University of Michigan - Ann Arbor

##### Jared E.

University of Winnipeg

### Video Transcript

in this problem on the topic of electric forces and fields were given three charges in arrangement as shown in the figure. We want to find the magnitude and direction of the electrostatic force that acts on the charge at the origin now in the sketch that we have drawn, We Can CFR as a result in force of the forces F six and F three that exerted on the charge of the region by the Three Newton column that the Nano column in the sixth column charges, respectively. So let's first calculate F six the force that the six new Nano column charge exerts on the charge of the Origen, which is the five knuckle um, charge. This is equal to K eight point 99 times 10 to the power nine Newton meter squared for Coolum squared times. The product of the charges Q one terms Q two, which is six times 10 to the minus nine Colon Times Q two, which is five times 10 to the minus nine columns divided by the separation between these charges squared which is zero 0.3 meters squared, so calculating we get F six to be three times 10 to the minus six Newton's or three Micro Newton's now the force that the minus three nana column charge exerts on the charges in the region. Similarly similarly, k 8.99 times 10 to the nine Newton Meter squared Cool um, squared times the product of the charges three times 10 to the minus nine columns multiplied by five times 10 to the minus nine columns divided by their separation, which is 0.1 m squared. So calculating we're getting magnitude of discharge on the charge of the region to be 1.35 times 10 to the minus five Newton's so we can now find the magnitude of the resultant fr. So if our is equal to the square root of F six squared plus F three squared and if we start do the values in here, we get this to be one 0.38 times 10 to the minus five Newtons. Now we need the angle theater and we know the Angle Theater is the Oct 10 of the vertical force F three of the horizontal of six. This gives us to to to be 77 0.5 degrees. So therefore, the resultant force. Acting at the original F R is equal to one 0.38 times 10 to the minus five Newton's, and it acts at an angle of 77.5 degrees below the negative X axis.

University of Kwazulu-Natal
##### Christina K.

Rutgers, The State University of New Jersey

##### Andy C.

University of Michigan - Ann Arbor

##### Jared E.

University of Winnipeg