In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge
E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis
E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis
The red arrow shows the direction of which the electric field points.
To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2
Etotal = sqrt(E1+E2) = 20494.97 V/m
E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2
d = 0.079 m = 7.9 cm
Answer:
Explanation:
Expression for time period of pendulum is given as follows
where l is length of pendulum and g is acceleration due to gravity .
Putting the given values for first place
Putting the values for second place
Dividing these two equation
T = 2.66455 s.
Answer:
<em>18808.7 m/s^2</em>
Explanation:
Given
Length of the pendulum L = 1.44 m
Number of complete cycles of oscillation n = 1.10 x 10^2
total time of oscillation t = 2.00 x 10^2 s
The period of the T = n/t
T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s
The period of a pendulum is gotten as
T =
where g is the acceleration due to gravity
substituting values, we have
0.55 =
0.0875 =
squaring both sides of the equation, we have
7.656 x 10^-3 = 144/g
g = 144/(7.656 x 10^-3) = <em>18808.7 m/s^2</em>