Answer:
Since the ball becomes positively charged, it will repel as like charges repel.
Answer:
The Magnitude of electric field is in the upward direction as shown directly towards the charge 
.
Explanation:
Given:
- side of a square,
- charge on one corner of the square,
- charge on the remaining 3 corners of the square,
<u>Distance of the center from each corners</u>
∴Distance of center from corners, 
Now, electric field due to charges is given as:
<u>For charge we have the field lines emerging out of the charge since it is positively charged:</u>
<u>Force by each of the charges at the remaining corners:</u>
<u> Now, net electric field in the vertical direction:</u>
<u>Now, net electric field in the horizontal direction:</u>
So the Magnitude of electric field is in the upward direction as shown directly towards the charge .
The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
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Answer:
1.23 m/s
Explanation:
p=mv
57.2 = 46.5v
v= 57.2/46.5
v= 1.23
If you want to verify your answer, just insert the value of v in the equation.
