Answer: 2.4×10^-3 v/m
Explanation: distance between plates of capacitor (d) =5.0×10^-3m
Potential difference between plates (v) = 12v
Force on electronic charge (f) = 3.8×10^-16 N
Strength of electric field (E) =?
The formulae that relates potential difference, eoectiic field strength and distance between plates is given as
v = Ed
By substituting the parameters, we have that
12 = E × 5.0×10^-3
E = 12/ 5.0 × 10^-3
E = 2.4×10^-3 v/m
Answer:
0.235 nC
Explanation:
Given:
= the magnitude of electric field = 
= the magnitude of electric force on each antenna = 
= The magnitude of charge on each antenna
Since the electric field is the electric force applied on a charged body of unit charge.
Hence, the value of q is 0.235 nC.
ELECTROSTATIC:
relating to stationary electric charges or fields as opposed to electric currents.
NEUTRAL:
nor negative nor positive/having no charge
POSITIVELY CHARGED:
positive charge occurs when the number of protons exceeds the number of electrons
NEGATIVELY CHARGED:
negative charge occurs when the number of electrons exceeds the number of protons.
COULOMB:
SI unit for electric charge. One coulomb is equal to the amount of charge from a current of one ampere flowing for one second.
MICROCOULOMB:
a unit of electrical charge equal to one millionth of a coulomb.
NANOCOULOMB:
Nanocoulombs are a unit of charge 1,000,000,000 times smaller than Coulomb.
CONSERVATION OF CHARGE:
constancy of the total electric charge in the universe or in any specific chemical or nuclear reaction
QUANTISATION OF CHARGE:
Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.
Answer:
The initial velocity was U=22.14m/s
Explanation:
Step one :
Applying the third equation of motion
v² = u²+ 2as
Where v= Final velocity
U =initial velocity
a= acceleration due to gravity
S= distance or displacement
Step two :
V= 0
a= 9.81m/s²
S=25m
U=?
Step three :
Substituting into the equation we have
0²=U²+2*9.81*25
0=U²+490.5
U²=-490.5
U=√490.5
U=22.14m/s
