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Electric field strength: e=f q
Point charge electric field strength: e=kq r
Uniform electric field: e=u d
Electric potential energy: e = q
Potential difference: u = -
Work done by electrostatic force: w = qu
Capacitance definition: c=q u
Capacitance: c = s 4 kd
The motion of charged particles in a uniform electric field.
Accelerated uniform electric field: 1 2*mv = qu
v² =2qu/m
Deflection of uniform intensity electric field:
Vertical acceleration: a=qu md
Vertical displacement: y=1 2*at =1 2*(qu md)*(x v).
Deflection angle: =v v =qux md(v).
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Commonly used formulas in electromagnetism.
Electric field strength: e=f q
Point charge electric field strength: e=kq r
Uniform electric field: e=u d
Electric potential energy: e = q
Potential difference: u = -
Work done by electrostatic force: w = qu
Capacitance definition: c=q u
Capacitance: c = s 4 kd
The motion of charged particles in a uniform electric field.
Accelerated uniform electric field: 1 2*mv = qu
v² =2qu/m
Deflection of uniform intensity electric field:
Vertical acceleration: a=qu md
Vertical displacement: y=1 2*at =1 2*(qu md)*(x v).
Deflection angle: =v v =qux md(v) micro-current: i=nesv
Non-electrostatic work of power supply: w= q
Ohm's law: i = u r
Series circuit Current: i = i = i = ......Voltage: u = u + u + u +
Parallel circuit voltage: u = u =u = ......
Current: i = i + i + i +
Resistors in series: r = r + r + r +
Resistors in parallel: 1 r = 1 r + 1 r + 1 r + Joule's law: q = i rt
p=i² r
p=u² /r
Electrical power: w=uit
Electric work: p=ui
Law of resistance: r = l s
Ohm's law for all circuits: =i(r+r).
U outside + U inside.
Ampere: f=ilbsin
Magnetic flux: =bs
Electromagnetic induction Induced electromotive force: e=nδ δt
Wire cutting magnetic inductance line: δs=lvδt
e=blv*sinθ
Induced electromotive force: e=lδi δt
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I can't show your problem here!
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Commonly used formulas in electromagnetism.
Electric field strength: e=f q
Point charge electric field strength: e=kq r2
Uniform electric field: e=u d
Electric potential energy: e? =qφ
Potential difference: u = - ?
Electrostatic force work: w??=qu??
Capacitance definition: c=q u
Capacitance: c = s 4 kd
The motion of charged particles in a uniform electric field.
Accelerated uniform electric field: 1 2 * mv2 = qu
v2 =2qu/m
Deflection of uniform intensity electric field:
Vertical acceleration: a=qu md
Vertical displacement: y=1 2*at? =1/2*(qu/md)*(x/v?)2
Deflection angle: =v v?=qux/md(v?2 Microscopic current: i=nesv
Non-electrostatic work of power supply: w= q
Ohm's law: i = u r
Series circuit current: i? =i? =i? =
Voltage: u = u? +u? +u? +
Parallel circuit voltage: u?=u?=u?=
Current: i = i?+i?+i?+
Resistance in series: r = r?+r?+r?+
Resistors in parallel: 1 r = 1 r?+1/r?+1/r?+ Joule's law: Q=I2 RT
p=i2 r
p=u2 /r
Electrical power: w=uit
Electric work: p=ui
Law of resistance: r = l s
Ohm's law for all circuits: =i(r+r).
U outside + U inside.
Ampere: f=ilbsin
Magnetic flux: =bs
Electromagnetic induction Induced electromotive force: e=nδ δt
Wire cutting magnetic inductance line: δs=lvδt
e=blv*sinθ
Induced electromotive force: e=lδi δt
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Electric field strength: e=f q
Point charge electric field strength: e=kq r
Uniform electric field: e=u d
Electric potential energy: e = q
Potential difference: u = -
Work done by electrostatic force: w = qu
Capacitance definition: c=q u
Capacitance: c = s 4 kd
The motion of charged particles in a uniform electric field.
Accelerated uniform electric field: 1 2*mv = qu
v² =2qu/m
Deflection of uniform intensity electric field:
Vertical acceleration: a=qu md
Vertical displacement: y=1 2*at =1 2*(qu md)*(x v).
Deflection angle: =v v =qux md(v).
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Physics is actually not difficult, the most important thing is whether the formula can be used and whether the analysis is in place. As long as you do more questions and learn to analyze more, you will find physics easy.