United States presidential election in New Jersey, 1860

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United States presidential election in New Jersey, 1860
New Jersey
← 1856 November 6, 1860 1864 →
  Abraham Lincoln by Alexander Helser, 1860-crop.jpg StephenADouglas.png
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel Vespasian Johnson
Electoral vote 4 3
Popular vote 58,346 62,869
Percentage 48.13%' 51.87%

President before election

James Buchanan

Elected President

Abraham Lincoln

The 1860 United States presidential election in New Jersey took place on November 6, 1860, as part of the 1860 United States presidential election. Voters chose seven electors of the Electoral College, who voted for president and vice president.

The winner in New Jersey depends on the votes. For the electoral votes, it was split, with Republican candidate Abraham Lincoln carrying four electoral votes, while Democratic candidate Stephen A. Douglas carried three. In the popular vote, Douglas won it by a narrow margin of 3.74%.


United States presidential election in New Jersey, 1860[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Abraham Lincoln of Illinois Hannibal Hamlin of Maine 58,346 48.13% 4 57.12%
Democratic Stephen A. Douglas of Illinois Herschel Vespasian Johnson of Georgia 62,869 51.87% 3 42.88%
Total 121,215 100.00% 7 100.00%


  1. ^ "1860 Presidential General Election Results - New Jersey". 
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