The electric field at the point x=5.00cm and y=0 points in the positive x direction with a magnitude of 10.0 n/c . at the point x=10.0cm and y=0 the electric field points in the positive x direction with a magnitude of 17.0 n/c . assume this electric field is produced by a single point charge. part afind the charge's location. part bfind the magnitude of the charge.

(A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.

Explanation:

Given that,

Distance in x axis = 5.00 cm

Electric field = 10.0 N/C

Distance in x axis = 10.0 cm

Electric field = 17.0 N/C

Since, q is the same charge, two formulas can be set equal using the two different electric fields.

(a). We need to calculate the location of the charge

Using formula of force

....(I)

Using formula of electric force

....(II)

From equation (I) and (II)

...(III)

For both points,

Put the value into the formula

Take the square root of both sides

(B). We need to calculate the charge

Using equation (III)

Put the value into the formula

Hence, (A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

not moving. pos accel,