The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
<h3>Electric field on the master charge</h3>
E = kq/r²
where;
- q is magnitude of master charge
- r is distance of separation
- k is Coulomb's constant
E = (9 x 10⁹ x 0.63)/(0.75²)
E = 1.008 x 10¹⁰ N/C
<h3>Force on the test charge</h3>
F = Eq
where;
- E is electric field
- q is the test charge
F = (1.008 x 10¹⁰) x (0.5)
F = 5.04 x 10⁹ N
Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
Learn more about electric field here: brainly.com/question/14372859
#SPJ1
Answer: 3 A
Explanation:
According to<u> Ohm's law</u>:
Where:
is the voltage
is the resistance of the resistor
is the electric current (the value we want to find)
Isolating :
Finally:
Ng seismic and translational waves we get the law of michio kaku.
Answer:
230 m/s northeast, 1.8 m/s up
Explanation:
204 kilometres = 204000 metres
15.0 minutes = 900 seconds
Velocity = Distance / Time
= 204000 / 900
= 230 m/s northeast (to 2 sf.)
1.6km = 1600 metres
Velocity = 1600 / 900
= 1.8 m/s up (to 2 sf.)
Read more on Brainly.com - brainly.com/question/13863590#readmore
